Time Limit: 5000MS | Memory Limit: 131072K | |
Total Submissions: 128810 | Accepted: 39971 | |
Case Time Limit: 2000MS |
Description
You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.
Input
The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A1, A2, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of Aa, Aa+1, ... , Ab.
Output
You need to answer all Q commands in order. One answer in a line.
Sample Input
10 5 1 2 3 4 5 6 7 8 9 10 Q 4 4 Q 1 10 Q 2 4 C 3 6 3 Q 2 4
Sample Output
4 55 9 15
Hint
Source
题目链接:http://poj.org/problem?id=3468
这道题是一个练习线段树区间更新很好的一个裸题;
直接一个Lazy_Tag模板就解决了
(PushUp(由下往上更新修改后的数据),Build(初始化线段树),Update(修改叶子结点的数据),Query(查询区间数据)
涉及到的位运算:n<<1 == n*2; n<<2 == n*4; n>>1==n/2; n<<1|1 == n*2+1;
AC代码如下:
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long LL;
const int maxn=1e5+10;
LL sum[maxn<<2];
LL cnt[maxn<<2]; ///延迟标记
int N,Q;
void build(int l,int r,int rt)///建树+结点lazy初始化
{
cnt[rt]=0;///lazy初始化
if(l==r)
{
scanf("%lld",&sum[rt]); ///输入时注意lld,没注意到WA了一次
return ;
}
int m=(l+r)>>1;///中间
build(l,m,rt<<1);///左子树
build(m+1,r,rt<<1|1);///右子树
sum[rt]=sum[rt<<1]+sum[rt<<1|1];///线段树功能:区间求和
}
void Down(int rt,int len)///查询lazy_Tag
{
if(cnt[rt])///存在标记,则向下传递标记
{///并更新左右子树的值
cnt[rt<<1]+=cnt[rt];///左子树
cnt[rt<<1|1]+=cnt[rt];///右子树
sum[rt<<1]+=cnt[rt]*(len-(len>>1));///更新左子树区间和
sum[rt<<1|1]+=cnt[rt]*(len>>1);///更新右子树区间和
cnt[rt]=0;///取消标记
}
}
LL Query(int a,int b,int l,int r,int rt)///查询区间和
{
if(a<=l && b>=r)
{
return sum[rt];
}
Down(rt,r-l+1);///查询是否需要更新值
///更新该结点的左右子树
int m=(l+r)>>1;
LL ans=0;
if(a<=m)
ans+=Query(a,b,l,m,rt<<1);
if(b>m)
ans+=Query(a,b,m+1,r,rt<<1|1);
///区间求和,所以ans+=Query(a,b,m+1,r,rt<<1|1)
///如果是求区间max则ans=max(ans,Query(a,b,m+1,r,rt<<1|1));
sum[rt]=sum[rt<<1]+sum[rt<<1|1];
return ans;
}
void Update(int a,int b,int c,int l,int r,int rt)
{///修改叶子结点的数据
if(a<=l && b>=r)
{
cnt[rt]+=c;///标记存在延迟修改量为C
sum[rt]+=(LL)(r-l+1)*c;///单点更新
///涉及*/运算时记得加强制转换
return ;
}
Down(rt,r-l+1);///更新该结点的左右子树的值
int m=(r+l)>>1;
if(a<=m)///更新左子树
Update(a,b,c,l,m,rt<<1);
if(b>m)///更新右子树
Update(a,b,c,m+1,r,rt<<1|1);
sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
int main()
{
while(scanf("%d%d",&N,&Q)==2)
{
build(1,N,1);
for(int i=0; i<Q; i++)
{
char ch;
int a,b,c;
getchar();
scanf("%c",&ch);
if(ch=='Q')
{
scanf("%d%d",&a,&b);
LL ans=Query(a,b,1,N,1);///查询区间数据
printf("%lld\n",ans);
}
else
{
scanf("%d%d%d",&a,&b,&c);
Update(a,b,c,1,N,1);///更新区间数据
}
}
}
return 0;
}