线段树汇总 之 练习篇-转自神牛notonlysuccess

FROM：http://www.notonlysuccess.com/index.php/segment-tree-complete/

在代码前先介绍一些我的线段树风格:

• maxn是题目给的最大区间,而节点数要开4倍,确切的来说节点数要开大于maxn的最小2^x的两倍
• lson和rson分辨表示结点的左儿子和右儿子,由于每次传参数的时候都固定是这几个变量,所以可以用预定于比较方便的表示
• 以前的写法是另外开两个个数组记录每个结点所表示的区间,其实这个区间不必保存,一边算一边传下去就行,只需要写函数的时候多两个参数,结合lson和rson的预定义可以很方便
• PushUP(int rt)是把当前结点的信息更新到父结点
• PushDown(int rt)是把当前结点的信息更新给儿子结点
• rt表示当前子树的根(root),也就是当前所在的结点

整理这些题目后我觉得线段树的题目整体上可以分成以下四个部分:

• 单点更新:最最基础的线段树,只更新叶子节点,然后把信息用PushUP(int r)这个函数更新上来
  • hdu1166 敌兵布阵
    题意:O(-1)
    思路:O(-1)
    线段树功能:update:单点增减 query:区间求和

    
           
           
    #include <cstdio>
    
    #define lson l , m , rt << 1
    #define rson m + 1 , r , rt << 1 | 1
    const int maxn = 55555;
    int sum[maxn<< 2];
    void PushUP(int rt) {
    sum[rt] = sum[rt<< 1] + sum[rt<< 1| 1];
    }
    void build(int l,int r,int rt) {
    if (l == r) {
    scanf( "%d",&sum[rt]);
    return ;
    }
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
    PushUP(rt);
    }
    void update(int p,int add,int l,int r,int rt) {
    if (l == r) {
    sum[rt] += add;
    return ;
    }
    int m = (l + r) >> 1;
    if (p <= m) update(p , add , lson);
    else update(p , add , rson);
    PushUP(rt);
    }
    int query(int L,int R,int l,int r,int rt) {
    if (L <= l && r <= R) {
    return sum[rt];
    }
    int m = (l + r) >> 1;
    int ret = 0;
    if (L <= m) ret += query(L , R , lson);
    if (R > m) ret += query(L , R , rson);
    return ret;
    }
    int main() {
    int T , n;
    scanf( "%d",&T);
    for ( int cas = 1 ; cas <= T ; cas ++) {
    printf( "Case %d:\n",cas);
    scanf( "%d",&n);
    build( 1 , n , 1);
    char op[ 10];
    while ( scanf( "%s",op)) {
    if (op[ 0] == 'E') break;
    int a , b;
    scanf( "%d%d",&a,&b);
    if (op[ 0] == 'Q') printf( "%d\n",query(a , b , 1 , n , 1));
    else if (op[ 0] == 'S') update(a , -b , 1 , n , 1);
    else update(a , b , 1 , n , 1);
    }
    }
    return 0;
    }
           
           
    

    
    
  • hdu1754 I Hate It
    题意:O(-1)
    思路:O(-1)
    线段树功能:update:单点替换 query:区间最值

    
           
           
    #include <cstdio>
    #include <algorithm>
    using namespace std;
    
    #define lson l , m , rt << 1
    #define rson m + 1 , r , rt << 1 | 1
    const int maxn = 222222;
    int MAX[maxn<< 2];
    void PushUP(int rt) {
    MAX[rt] = max(MAX[rt<< 1] , MAX[rt<< 1| 1]);
    }
    void build(int l,int r,int rt) {
    if (l == r) {
    scanf( "%d",&MAX[rt]);
    return ;
    }
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
    PushUP(rt);
    }
    void update(int p,int sc,int l,int r,int rt) {
    if (l == r) {
    MAX[rt] = sc;
    return ;
    }
    int m = (l + r) >> 1;
    if (p <= m) update(p , sc , lson);
    else update(p , sc , rson);
    PushUP(rt);
    }
    int query(int L,int R,int l,int r,int rt) {
    if (L <= l && r <= R) {
    return MAX[rt];
    }
    int m = (l + r) >> 1;
    int ret = 0;
    if (L <= m) ret = max(ret , query(L , R , lson));
    if (R > m) ret = max(ret , query(L , R , rson));
    return ret;
    }
    int main() {
    int n , m;
    while (~ scanf( "%d%d",&n,&m)) {
    build( 1 , n , 1);
    while (m --) {
    char op[ 2];
    int a , b;
    scanf( "%s%d%d",op,&a,&b);
    if (op[ 0] == 'Q') printf( "%d\n",query(a , b , 1 , n , 1));
    else update(a , b , 1 , n , 1);
    }
    }
    return 0;
    }
           
           
    

    
    
  • hdu1394 Minimum Inversion Number
    题意:求Inversion后的最小逆序数
    思路:用O(nlogn)复杂度求出最初逆序数后,就可以用O(1)的复杂度分别递推出其他解
    线段树功能:update:单点增减 query:区间求和

    
           
           
    #include <cstdio>
    #include <algorithm>
    using namespace std;
    
    #define lson l , m , rt << 1
    #define rson m + 1 , r , rt << 1 | 1
    const int maxn = 5555;
    int sum[maxn<< 2];
    void PushUP(int rt) {
    sum[rt] = sum[rt<< 1] + sum[rt<< 1| 1];
    }
    void build(int l,int r,int rt) {
    sum[rt] = 0;
    if (l == r) return ;
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
    }
    void update(int p,int l,int r,int rt) {
    if (l == r) {
    sum[rt] ++;
    return ;
    }
    int m = (l + r) >> 1;
    if (p <= m) update(p , lson);
    else update(p , rson);
    PushUP(rt);
    }
    int query(int L,int R,int l,int r,int rt) {
    if (L <= l && r <= R) {
    return sum[rt];
    }
    int m = (l + r) >> 1;
    int ret = 0;
    if (L <= m) ret += query(L , R , lson);
    if (R > m) ret += query(L , R , rson);
    return ret;
    }
    int x[maxn];
    int main() {
    int n;
    while (~ scanf( "%d",&n)) {
    build( 0 , n - 1 , 1);
    int sum = 0;
    for ( int i = 0 ; i < n ; i ++) {
    scanf( "%d",&x[i]);
    sum += query(x[i] , n - 1 , 0 , n - 1 , 1);
    update(x[i] , 0 , n - 1 , 1);
    }
    int ret = sum;
    for ( int i = 0 ; i < n ; i ++) {
    sum += n - x[i] - x[i] - 1;
    ret = min(ret , sum);
    }
    printf( "%d\n",ret);
    }
    return 0;
    }
           
           
    

    
    
  • hdu2795 Billboard
    题意:h*w的木板,放进一些1*L的物品,求每次放空间能容纳且最上边的位子
    思路:每次找到最大值的位子,然后减去L
    线段树功能:query:区间求最大值的位子(直接把update的操作在query里做了)

    
           
           
    #include <cstdio>
    #include <algorithm>
    using namespace std;
    
    #define lson l , m , rt << 1
    #define rson m + 1 , r , rt << 1 | 1
    const int maxn = 222222;
    int h , w , n;
    int MAX[maxn<< 2];
    void PushUP(int rt) {
    MAX[rt] = max(MAX[rt<< 1] , MAX[rt<< 1| 1]);
    }
    void build(int l,int r,int rt) {
    MAX[rt] = w;
    if (l == r) return ;
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
    }
    int query(int x,int l,int r,int rt) {
    if (l == r) {
    MAX[rt] -= x;
    return l;
    }
    int m = (l + r) >> 1;
    int ret = (MAX[rt<< 1] >= x) ? query(x , lson) : query(x , rson);
    PushUP(rt);
    return ret;
    }
    int main() {
    while (~ scanf( "%d%d%d",&h,&w,&n)) {
    if (h > n) h = n;
    build( 1 , h , 1);
    while (n --) {
    int x;
    scanf( "%d",&x);
    if (MAX[ 1] < x) puts( "-1");
    else printf( "%d\n",query(x , 1 , h , 1));
    }
    }
    return 0;
    }
           
           
    

    练习：
  • poj2828 Buy Tickets
    poj2886 Who Gets the Most Candies?
• 成段更新(通常这对初学者来说是一道坎),需要用到延迟标记(或者说懒惰标记),简单来说就是每次更新的时候不要更新到底,用延迟标记使得更新延迟到下次需要更新or询问到的时候
  • hdu1698 Just a Hook
    题意:O(-1)
    思路:O(-1)
    线段树功能:update:成段替换 (由于只query一次总区间,所以可以直接输出1结点的信息)

    
           
           
    #include <cstdio>
    #include <algorithm>
    using namespace std;
    
    #define lson l , m , rt << 1
    #define rson m + 1 , r , rt << 1 | 1
    const int maxn = 111111;
    int h , w , n;
    int col[maxn<< 2];
    int sum[maxn<< 2];
    void PushUp(int rt) {
    sum[rt] = sum[rt<< 1] + sum[rt<< 1| 1];
    }
    void PushDown(int rt,int m) {
    if (col[rt]) {
    col[rt<< 1] = col[rt<< 1| 1] = col[rt];
    sum[rt<< 1] = (m - (m >> 1)) * col[rt];
    sum[rt<< 1| 1] = (m >> 1) * col[rt];
    col[rt] = 0;
    }
    }
    void build(int l,int r,int rt) {
    col[rt] = 0;
    sum[rt] = 1;
    if (l == r) return ;
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
    PushUp(rt);
    }
    void update(int L,int R,int c,int l,int r,int rt) {
    if (L <= l && r <= R) {
    col[rt] = c;
    sum[rt] = c * (r - l + 1);
    return ;
    }
    PushDown(rt , r - l + 1);
    int m = (l + r) >> 1;
    if (L <= m) update(L , R , c , lson);
    if (R > m) update(L , R , c , rson);
    PushUp(rt);
    }
    int main() {
    int T , n , m;
    scanf( "%d",&T);
    for ( int cas = 1 ; cas <= T ; cas ++) {
    scanf( "%d%d",&n,&m);
    build( 1 , n , 1);
    while (m --) {
    int a , b , c;
    scanf( "%d%d%d",&a,&b,&c);
    update(a , b , c , 1 , n , 1);
    }
    printf( "Case %d: The total value of the hook is %d.\n",cas , sum[ 1]);
    }
    return 0;
    }
           
           
    

    
    
  • poj3468 A Simple Problem with Integers
  • 题意:O(-1)
    思路:O(-1)
    线段树功能:update:成段增减 query:区间求和

    
           
           
    #include <cstdio>
    #include <algorithm>
    using namespace std;
    
    #define lson l , m , rt << 1
    #define rson m + 1 , r , rt << 1 | 1
    #define LL long long
    const int maxn = 111111;
    LL add[maxn<< 2];
    LL sum[maxn<< 2];
    void PushUp(int rt) {
    sum[rt] = sum[rt<< 1] + sum[rt<< 1| 1];
    }
    void PushDown(int rt,int m) {
    if (add[rt]) {
    add[rt<< 1] += add[rt];
    add[rt<< 1| 1] += add[rt];
    sum[rt<< 1] += add[rt] * (m - (m >> 1));
    sum[rt<< 1| 1] += add[rt] * (m >> 1);
    add[rt] = 0;
    }
    }
    void build(int l,int r,int rt) {
    add[rt] = 0;
    if (l == r) {
    scanf( "%lld",&sum[rt]);
    return ;
    }
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
    PushUp(rt);
    }
    void update(int L,int R,int c,int l,int r,int rt) {
    if (L <= l && r <= R) {
    add[rt] += c;
    sum[rt] += (LL)c * (r - l + 1);
    return ;
    }
    PushDown(rt , r - l + 1);
    int m = (l + r) >> 1;
    if (L <= m) update(L , R , c , lson);
    if (m < R) update(L , R , c , rson);
    PushUp(rt);
    }
    LL query(int L,int R,int l,int r,int rt) {
    if (L <= l && r <= R) {
    return sum[rt];
    }
    PushDown(rt , r - l + 1);
    int m = (l + r) >> 1;
    LL ret = 0;
    if (L <= m) ret += query(L , R , lson);
    if (m < R) ret += query(L , R , rson);
    return ret;
    }
    int main() {
    int N , Q;
    scanf( "%d%d",&N,&Q);
    build( 1 , N , 1);
    while (Q --) {
    char op[ 2];
    int a , b , c;
    scanf( "%s",op);
    if (op[ 0] == 'Q') {
    scanf( "%d%d",&a,&b);
    printf( "%lld\n",query(a , b , 1 , N , 1));
    } else {
    scanf( "%d%d%d",&a,&b,&c);
    update(a , b , c , 1 , N , 1);
    }
    }
    return 0;
    }
           
           
    

    
    
  • poj2528 Mayor’s posters
  • 题意:在墙上贴海报,海报可以互相覆盖,问最后可以看见几张海报
    思路:这题数据范围很大,直接搞超时+超内存,需要离散化:
    离散化简单的来说就是只取我们需要的值来用,比如说区间[1000,2000],[1990,2012] 我们用不到[-∞,999][1001,1989][1991,1999][2001,2011][2013,+∞]这些值,所以我只需要1000,1990,2000,2012就够了,将其分别映射到0,1,2,3,在于复杂度就大大的降下来了
    所以离散化要保存所有需要用到的值,排序后,分别映射到1~n,这样复杂度就会小很多很多
    而这题的难点在于每个数字其实表示的是一个单位长度(并非一个点),这样普通的离散化会造成许多错误(包括我以前的代码,poj这题数据奇弱)
    给出下面两个简单的例子应该能体现普通离散化的缺陷:
    例子一:1-10 1-4 5-10
    例子二:1-10 1-4 6-10
    普通离散化后都变成了[1,4][1,2][3,4]
    线段2覆盖了[1,2],线段3覆盖了[3,4],那么线段1是否被完全覆盖掉了呢?
    例子一是完全被覆盖掉了,而例子二没有被覆盖

    为了解决这种缺陷,我们可以在排序后的数组上加些处理,比如说[1,2,6,10]
    如果相邻数字间距大于1的话,在其中加上任意一个数字,比如加成[1,2,3,6,7,10],然后再做线段树就好了.
    线段树功能:update:成段替换 query:简单hash

    
           
           
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    using namespace std;
    #define lson l , m , rt << 1
    #define rson m + 1 , r , rt << 1 | 1
    
    const int maxn = 11111;
    bool hash[maxn];
    int li[maxn] , ri[maxn];
    int X[maxn* 3];
    int col[maxn<< 4];
    int cnt;
    
    void PushDown(int rt) {
    if (col[rt] != -1) {
    col[rt<< 1] = col[rt<< 1| 1] = col[rt];
    col[rt] = -1;
    }
    }
    void update(int L,int R,int c,int l,int r,int rt) {
    if (L <= l && r <= R) {
    col[rt] = c;
    return ;
    }
    PushDown(rt);
    int m = (l + r) >> 1;
    if (L <= m) update(L , R , c , lson);
    if (m < R) update(L , R , c , rson);
    }
    void query(int l,int r,int rt) {
    if (col[rt] != -1) {
    if (!hash[col[rt]]) cnt ++;
    hash[ col[rt] ] = true;
    return ;
    }
    if (l == r) return ;
    int m = (l + r) >> 1;
    query(lson);
    query(rson);
    }
    int Bin(int key,int n,int X[]) {
    int l = 0 , r = n - 1;
    while (l <= r) {
    int m = (l + r) >> 1;
    if (X[m] == key) return m;
    if (X[m] < key) l = m + 1;
    else r = m - 1;
    }
    return -1;
    }
    int main() {
    int T , n;
    scanf( "%d",&T);
    while (T --) {
    scanf( "%d",&n);
    int nn = 0;
    for ( int i = 0 ; i < n ; i ++) {
    scanf( "%d%d",&li[i] , &ri[i]);
    X[nn++] = li[i];
    X[nn++] = ri[i];
    }
    sort(X , X + nn);
    int m = 1;
    for ( int i = 1 ; i < nn; i ++) {
    if (X[i] != X[i -1]) X[m ++] = X[i];
    }
    for ( int i = m - 1 ; i > 0 ; i --) {
    if (X[i] != X[i -1] + 1) X[m ++] = X[i -1] + 1;
    }
    sort(X , X + m);
    memset(col , -1 , sizeof(col));
    for ( int i = 0 ; i < n ; i ++) {
    int l = Bin(li[i] , m , X);
    int r = Bin(ri[i] , m , X);
    update(l , r , i , 0 , m , 1);
    }
    cnt = 0;
    memset(hash , false , sizeof(hash));
    query( 0 , m , 1);
    printf( "%d\n",cnt);
    }
    return 0;
    }
           
           
    

    
    

  • poj3225 Help with Intervals
    题意:区间操作,交,并,补等
    思路:
    我们一个一个操作来分析:(用0和1表示是否包含区间,-1表示该区间内既有包含又有不包含)
    U:把区间[l,r]覆盖成1
    I:把[-∞,l)(r,∞]覆盖成0
    D:把区间[l,r]覆盖成0
    C:把[-∞,l)(r,∞]覆盖成0 , 且[l,r]区间0/1互换
    S:[l,r]区间0/1互换

    成段覆盖的操作很简单,比较特殊的就是区间0/1互换这个操作,我们可以称之为异或操作
    很明显我们可以知道这个性质:当一个区间被覆盖后,不管之前有没有异或标记都没有意义了
    所以当一个节点得到覆盖标记时把异或标记清空
    而当一个节点得到异或标记的时候,先判断覆盖标记,如果是0或1,直接改变一下覆盖标记,不然的话改变异或标记

    开区间闭区间只要数字乘以2就可以处理(偶数表示端点,奇数表示两端点间的区间)
    线段树功能:update:成段替换,区间异或 query:简单hash

    
           
           
    #include <cstdio>
    #include <cstring>
    #include <cctype>
    #include <algorithm>
    using namespace std;
    #define lson l , m , rt << 1
    #define rson m + 1 , r , rt << 1 | 1
    
    const int maxn = 131072;
    bool hash[maxn+ 1];
    int cover[maxn<< 2];
    int XOR[maxn<< 2];
    void FXOR(int rt) {
    if (cover[rt] != -1) cover[rt] ^= 1;
    else XOR[rt] ^= 1;
    }
    void PushDown(int rt) {
    if (cover[rt] != -1) {
    cover[rt<< 1] = cover[rt<< 1| 1] = cover[rt];
    XOR[rt<< 1] = XOR[rt<< 1| 1] = 0;
    cover[rt] = -1;
    }
    if (XOR[rt]) {
    FXOR(rt<< 1);
    FXOR(rt<< 1| 1);
    XOR[rt] = 0;
    }
    }
    void update(char op,int L,int R,int l,int r,int rt) {
    if (L <= l && r <= R) {
    if (op == 'U') {
    cover[rt] = 1;
    XOR[rt] = 0;
    } else if (op == 'D') {
    cover[rt] = 0;
    XOR[rt] = 0;
    } else if (op == 'C' || op == 'S') {
    FXOR(rt);
    }
    return ;
    }
    PushDown(rt);
    int m = (l + r) >> 1;
    if (L <= m) update(op , L , R , lson);
    else if (op == 'I' || op == 'C') {
    XOR[rt<< 1] = cover[rt<< 1] = 0;
    }
    if (m < R) update(op , L , R , rson);
    else if (op == 'I' || op == 'C') {
    XOR[rt<< 1| 1] = cover[rt<< 1| 1] = 0;
    }
    }
    void query(int l,int r,int rt) {
    if (cover[rt] == 1) {
    for ( int it = l ; it <= r ; it ++) {
    hash[it] = true;
    }
    return ;
    } else if (cover[rt] == 0) return ;
    if (l == r) return ;
    PushDown(rt);
    int m = (l + r) >> 1;
    query(lson);
    query(rson);
    }
    int main() {
    cover[ 1] = XOR[ 1] = 0;
    char op , l , r;
    int a , b;
    while ( ~ scanf( "%c %c%d,%d%c\n",&op , &l , &a , &b , &r) ) {
    a <<= 1 , b <<= 1;
    if (l == '(') a ++;
    if (r == ')') b --;
    if (a > b) {
    if (op == 'C' || op == 'I') {
    cover[ 1] = XOR[ 1] = 0;
    }
    } else update(op , a , b , 0 , maxn , 1);
    }
    query( 0 , maxn , 1);
    bool flag = false;
    int s = -1 , e;
    for ( int i = 0 ; i <= maxn ; i ++) {
    if (hash[i]) {
    if (s == -1) s = i;
    e = i;
    } else {
    if (s != -1) {
    if (flag) printf( " ");
    flag = true;
    printf( "%c%d,%d%c",s& 1? '(': '[' , s>> 1 , (e+ 1)>> 1 , e& 1? ')': ']');
    s = -1;
    }
    }
    }
    if (!flag) printf( "empty set");
    puts( "");
    return 0;
    }
           
           
    

    
    

  • poj1436 Horizontally Visible Segments
    poj2991 Crane
    Another LCIS
    Bracket Sequence
• 区间合并
  这类题目会询问区间中满足条件的连续最长区间,所以PushUp的时候需要对左右儿子的区间进行合并
  • poj3667 Hotel
    题意:1 a:询问是不是有连续长度为a的空房间,有的话住进最左边
    2 a b:将[a,a+b-1]的房间清空
    思路:记录区间中最长的空房间
    线段树操作:update:区间替换 query:询问满足条件的最左断点

    
           
           
    #include <cstdio>
    #include <cstring>
    #include <cctype>
    #include <algorithm>
    using namespace std;
    #define lson l , m , rt << 1
    #define rson m + 1 , r , rt << 1 | 1
    
    const int maxn = 55555;
    int lsum[maxn<< 2] , rsum[maxn<< 2] , msum[maxn<< 2];
    int cover[maxn<< 2];
    
    void PushDown(int rt,int m) {
    if (cover[rt] != -1) {
    cover[rt<< 1] = cover[rt<< 1| 1] = cover[rt];
    msum[rt<< 1] = lsum[rt<< 1] = rsum[rt<< 1] = cover[rt] ? 0 : m - (m >> 1);
    msum[rt<< 1| 1] = lsum[rt<< 1| 1] = rsum[rt<< 1| 1] = cover[rt] ? 0 : (m >> 1);
    cover[rt] = -1;
    }
    }
    void PushUp(int rt,int m) {
    lsum[rt] = lsum[rt<< 1];
    rsum[rt] = rsum[rt<< 1| 1];
    if (lsum[rt] == m - (m >> 1)) lsum[rt] += lsum[rt<< 1| 1];
    if (rsum[rt] == (m >> 1)) rsum[rt] += rsum[rt<< 1];
    msum[rt] = max(lsum[rt<< 1| 1] + rsum[rt<< 1] , max(msum[rt<< 1] , msum[rt<< 1| 1]));
    }
    void build(int l,int r,int rt) {
    msum[rt] = lsum[rt] = rsum[rt] = r - l + 1;
    cover[rt] = -1;
    if (l == r) return ;
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
    }
    void update(int L,int R,int c,int l,int r,int rt) {
    if (L <= l && r <= R) {
    msum[rt] = lsum[rt] = rsum[rt] = c ? 0 : r - l + 1;
    cover[rt] = c;
    return ;
    }
    PushDown(rt , r - l + 1);
    int m = (l + r) >> 1;
    if (L <= m) update(L , R , c , lson);
    if (m < R) update(L , R , c , rson);
    PushUp(rt , r - l + 1);
    }
    int query(int w,int l,int r,int rt) {
    if (l == r) return l;
    PushDown(rt , r - l + 1);
    int m = (l + r) >> 1;
    if (msum[rt<< 1] >= w) return query(w , lson);
    else if (rsum[rt<< 1] + lsum[rt<< 1| 1] >= w) return m - rsum[rt<< 1] + 1;
    return query(w , rson);
    }
    int main() {
    int n , m;
    scanf( "%d%d",&n,&m);
    build( 1 , n , 1);
    while (m --) {
    int op , a , b;
    scanf( "%d",&op);
    if (op == 1) {
    scanf( "%d",&a);
    if (msum[ 1] < a) puts( "0");
    else {
    int p = query(a , 1 , n , 1);
    printf( "%d\n",p);
    update(p , p + a - 1 , 1 , 1 , n , 1);
    }
    } else {
    scanf( "%d%d",&a,&b);
    update(a , a + b - 1 , 0 , 1 , n , 1);
    }
    }
    return 0;
    }
           
           
    

    
    
  •  练习
    hdu3308 LCIS
    hdu3397 Sequence operation
    hdu2871 Memory Control
    hdu1540 Tunnel Warfare
    CF46-D Parking Lot
• 扫描线
  这类题目需要将一些操作排序,然后从左到右用一根扫描线(当然是在我们脑子里)扫过去
  最典型的就是矩形面积并,周长并等题
  • hdu1542 Atlantis
    题意:矩形面积并
    思路:浮点数先要离散化;然后把矩形分成两条边,上边和下边,对横轴建树,然后从下到上扫描上去,用cnt表示该区间下边比上边多几个,sum代表该区间内被覆盖的线段的长度总和
    这里线段树的一个结点并非是线段的一个端点,而是该端点和下一个端点间的线段,所以题目中r+1,r-1的地方可以自己好好的琢磨一下
    线段树操作:update:区间增减 query:直接取根节点的值

    
           
           
    #include <cstdio>
    #include <cstring>
    #include <cctype>
    #include <algorithm>
    using namespace std;
    #define lson l , m , rt << 1
    #define rson m + 1 , r , rt << 1 | 1
    
    const int maxn = 2222;
    int cnt[maxn << 2];
    double sum[maxn << 2];
    double X[maxn];
    struct Seg {
    double h , l , r;
    int s;
    Seg(){}
    Seg( double a, double b, double c, int d) : l(a) , r(b) , h(c) , s(d) {}
    bool operator < ( const Seg &cmp) const {
    return h < cmp.h;
    }
    }ss[maxn];
    void PushUp(int rt,int l,int r) {
    if (cnt[rt]) sum[rt] = X[r+ 1] - X[l];
    else if (l == r) sum[rt] = 0;
    else sum[rt] = sum[rt<< 1] + sum[rt<< 1| 1];
    }
    void update(int L,int R,int c,int l,int r,int rt) {
    if (L <= l && r <= R) {
    cnt[rt] += c;
    PushUp(rt , l , r);
    return ;
    }
    int m = (l + r) >> 1;
    if (L <= m) update(L , R , c , lson);
    if (m < R) update(L , R , c , rson);
    PushUp(rt , l , r);
    }
    int Bin(double key,int n,double X[]) {
    int l = 0 , r = n - 1;
    while (l <= r) {
    int m = (l + r) >> 1;
    if (X[m] == key) return m;
    if (X[m] < key) l = m + 1;
    else r = m - 1;
    }
    return -1;
    }
    int main() {
    int n , cas = 1;
    while (~ scanf( "%d",&n) && n) {
    int m = 0;
    while (n --) {
    double a , b , c , d;
    scanf( "%lf%lf%lf%lf",&a,&b,&c,&d);
    X[m] = a;
    ss[m++] = Seg(a , c , b , 1);
    X[m] = c;
    ss[m++] = Seg(a , c , d , -1);
    }
    sort(X , X + m);
    sort(ss , ss + m);
    int k = 1;
    for ( int i = 1 ; i < m ; i ++) {
    if (X[i] != X[i -1]) X[k++] = X[i];
    }
    memset(cnt , 0 , sizeof(cnt));
    memset(sum , 0 , sizeof(sum));
    double ret = 0;
    for ( int i = 0 ; i < m - 1 ; i ++) {
    int l = Bin(ss[i].l , k , X);
    int r = Bin(ss[i].r , k , X) - 1;
    if (l <= r) update(l , r , ss[i].s , 0 , k - 1, 1);
    ret += sum[ 1] * (ss[i+ 1].h - ss[i].h);
    }
    printf( "Test case #%d\nTotal explored area: %.2lf\n\n",cas++ , ret);
    }
    return 0;
    }
           
           
    

    
    
  • hdu1828 Picture
    题意:矩形周长并
    思路:与面积不同的地方是还要记录竖的边有几个(numseg记录),并且当边界重合的时候需要合并(用lbd和rbd表示边界来辅助)
    线段树操作:update:区间增减 query:直接取根节点的值

    
           
           
    #include <cstdio>
    #include <cstring>
    #include <cctype>
    #include <algorithm>
    using namespace std;
    #define lson l , m , rt << 1
    #define rson m + 1 , r , rt << 1 | 1
    
    const int maxn = 22222;
    struct Seg{
    int l , r , h , s;
    Seg() {}
    Seg( int a, int b, int c, int d):l(a) , r(b) , h(c) , s(d) {}
    bool operator < ( const Seg &cmp) const {
    if (h == cmp.h) return s > cmp.s;
    return h < cmp.h;
    }
    }ss[maxn];
    bool lbd[maxn<< 2] , rbd[maxn<< 2];
    int numseg[maxn<< 2];
    int cnt[maxn<< 2];
    int len[maxn<< 2];
    void PushUP(int rt,int l,int r) {
    if (cnt[rt]) {
    lbd[rt] = rbd[rt] = 1;
    len[rt] = r - l + 1;
    numseg[rt] = 2;
    } else if (l == r) {
    len[rt] = numseg[rt] = lbd[rt] = rbd[rt] = 0;
    } else {
    lbd[rt] = lbd[rt<< 1];
    rbd[rt] = rbd[rt<< 1| 1];
    len[rt] = len[rt<< 1] + len[rt<< 1| 1];
    numseg[rt] = numseg[rt<< 1] + numseg[rt<< 1| 1];
    if (lbd[rt<< 1| 1] && rbd[rt<< 1]) numseg[rt] -= 2; //两条线重合
    }
    }
    void update(int L,int R,int c,int l,int r,int rt) {
    if (L <= l && r <= R) {
    cnt[rt] += c;
    PushUP(rt , l , r);
    return ;
    }
    int m = (l + r) >> 1;
    if (L <= m) update(L , R , c , lson);
    if (m < R) update(L , R , c , rson);
    PushUP(rt , l , r);
    }
    int main() {
    int n;
    while (~ scanf( "%d",&n)) {
    int m = 0;
    int lbd = 10000, rbd = -10000;
    for ( int i = 0 ; i < n ; i ++) {
    int a , b , c , d;
    scanf( "%d%d%d%d",&a,&b,&c,&d);
    lbd = min(lbd , a);
    rbd = max(rbd , c);
    ss[m++] = Seg(a , c , b , 1);
    ss[m++] = Seg(a , c , d , -1);
    }
    sort(ss , ss + m);
    int ret = 0 , last = 0;
    for ( int i = 0 ; i < m ; i ++) {
    if (ss[i].l < ss[i].r) update(ss[i].l , ss[i].r - 1 , ss[i].s , lbd , rbd - 1 , 1);
    ret += numseg[ 1] * (ss[i+ 1].h - ss[i].h);
    ret += abs(len[ 1] - last);
    last = len[ 1];
    }
    printf( "%d\n",ret);
    }
    return 0;
    }
           
           
    

    
    
  • 练习
    hdu3265 Posters
    hdu3642 Get The Treasury
    poj2482 Stars in Your Window
    poj2464 Brownie Points II
    hdu3255 Farming 
    ural1707 Hypnotoad's Secret
    uva11983 Weird Advertisement

线段树与其他结合练习(欢迎大家补充):

• hdu3954 Level up
• hdu4027 Can you answer these queries?
• hdu3333 Turing Tree
• hdu3874 Necklace
• hdu3016 Man Down
• hdu3340 Rain in ACStar
• zju3511 Cake Robbery
• UESTC1558 Charitable Exchange
• CF85-D Sum of Medians
• spojGSS2 Can you answer these queries II