CodeForces 765 F Souvenirs

Souvenirs

题意：给你n个数， m次询问， 对于每次一次询问， 求出询问区间内绝对值差值的最小值。

题解：先按查询的右端点从小到大sort一下，然后对于塞入一个数的时候， 就处理出所有左端点到目前位置的点， 然后查询。

（待补，先上课。）

代码：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
 4 #define LL long long
 5 #define ULL unsigned LL
 6 #define fi first
 7 #define se second
 8 #define pb push_back
 9 #define lson l,m,rt<<1
10 #define rson m+1,r,rt<<1|1
11 #define max3(a,b,c) max(a,max(b,c))
12 #define min3(a,b,c) min(a,min(b,c))
13 typedef pair<int,int> pll;
14 const int INF = 0x3f3f3f3f;
15 const LL mod =  (int)1e9+7;
16 const int N = 1e5 + 100;
17 vector<int> vc[N<<2];
18 int Min[N<<2];
19 int a[N];
20 int n;
21 struct Node{
22     int l, r;
23     int id;
24 }q[N*3];
25 bool cmp(Node x1, Node x2){
26     return x1.r < x2.r;
27 }
28 void Merge(int rt){
29     int i = 0, j = 0;
30     int l = rt*2, r = rt*2+1;
31     int szl = vc[l].size(), szr = vc[r].size();
32     while(i < szl && j < szr){
33         if(vc[l][i] < vc[r][j]) vc[rt].pb(vc[l][i]), i++;
34         else vc[rt].pb(vc[r][j]), j++;
35     }
36     while(i < szl) vc[rt].pb(vc[l][i]), i++;
37     while(j < szr) vc[rt].pb(vc[r][j]), j++;
38 }
39 void Build(int l, int r, int rt){
40     Min[rt] = INF;
41     if(l == r){
42         vc[rt].pb(a[l]);
43         return ;
44     }
45     int m = l+r >> 1;
46     Build(lson);
47     Build(rson);
48     Merge(rt);
49 }
50 int Query(int l, int r, int rt, int L, int R){
51     if(L <= l && r <= R){
52         return Min[rt];
53     }
54     int ret = INF, m = l+r >> 1;
55     if(L <= m) ret = min(ret, Query(lson,L,R));
56     if(m < R) ret = min(ret, Query(rson,L,R));
57     return ret;
58 }
59 vector<int>::iterator it;
60 void Update(int l, int r, int rt, int R, int v, int &d){
61     if(l == r){
62         Min[rt] = min(Min[rt], abs(v-vc[rt][0]));
63         d = min(d, Min[rt]);
64         return ;
65     }
66     it = lower_bound(vc[rt].begin(), vc[rt].end(), v);
67     if((it == vc[rt].end() || *it-v >= d) && (it == vc[rt].begin() || v - *(--it) >= d)){
68         d = min(d, Query(1, n, 1, l, R));
69         return ;
70     }
71     int m = l+r >> 1;
72     if(R > m) Update(rson,R,v,d);
73     Update(lson,R,v,d);
74     Min[rt] = min3(Min[rt], Min[rt*2], Min[rt*2+1]);
75 }
76 int ans[N*3];
77 int main(){
78     scanf("%d", &n);
79     for(int i = 1; i <= n; i++)  scanf("%d", &a[i]);
80     Build(1,n,1);
81     int m;
82     scanf("%d", &m);
83     for(int i = 1; i <= m; i++)  scanf("%d%d", &q[i].l, &q[i].r), q[i].id = i;
84     sort(q+1, q+1+m, cmp);
85     int r = 1;
86     int d;
87     for(int i = 1; i <= m; i++){
88         while(r < q[i].r) {
89             d = INF;
90             Update(1, n, 1, r, a[r+1], d);
91             r++;
92         }
93         ans[ q[i].id ] = Query(1, n, 1, q[i].l, r);
94     }
95     for(int i = 1; i <= m; i++){
96         printf("%d\n", ans[i]);
97     }
98     return 0;
99 }

765F