CodeForces 669 E Little Artem and Time Machine CDQ分治

题目传送门

题意：现在有3种操作，

1 t x 在t秒往multiset里面插入一个x

2 t x 在t秒从multiset里面删除一个x

3 t x 在t秒查询multiset里面有多少x

事情是按照输入顺序发生的，这个人有一个时光机，可以穿梭到那一秒去执行操作。

题解：CDQ分治。3维偏序，第一维是输入顺序，第二维t，然后直接map处理数据就好了。

代码：

 #include<bits/stdc++.h>
 using namespace std;
 #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
 #define LL long long
 #define ULL unsigned LL
 #define fi first
 #define se second
 #define pb push_back
 #define lson l,m,rt<<1
 #define rson m+1,r,rt<<1|1
 #define lch(x) tr[x].son[0]
 #define rch(x) tr[x].son[1]
 #define max3(a,b,c) max(a,max(b,c))
 #define min3(a,b,c) min(a,min(b,c))
 typedef pair<int,int> pll;
 const int inf = 0x3f3f3f3f;
 const LL INF = 0x3f3f3f3f3f3f3f3f;
 const LL mod =  (int)1e9+;
 const int N = 1e5 + ;
 struct Node{
     int op, t, m, id;
     bool operator < (const Node & x){
         return t < x.t;
     }
 }A[N], tmp[N];
 int ans[N];
 map<int,int> mp;
 void cdq(int l, int r){
     if(l >= r) return ;
     int mid = l + r >> ;
     cdq(l, mid);
     cdq(mid+, r);
     int top = , tl = l, tr = mid + ;
     while(tl <= mid && tr <= r){
         if(A[tl].t <= A[tr].t){
             tmp[++top] = A[tl++];
         }
         else tmp[++top] = A[tr++];
     }
     while(tl <= mid){
         tmp[++top] = A[tl++];
     }
     while(tr <= r){
         tmp[++top] = A[tr++];
     }
     for(int i = ; i <= top; i++){
         int id = tmp[i].id, op = tmp[i].op;
         int m = tmp[i].m;
         if(id <= mid && op != ){
             if(op == ) mp[m]++;
             else mp[m]--;
         }
         else if(id > mid && op == ){
             ans[id] += mp[m];
         }
         A[l+i-] = tmp[i];
     }
     for(int i = ; i <= top; i++){
         int id = tmp[i].id, op = tmp[i].op;
         int m = tmp[i].m;
         if(id <= mid && op != ){
             if(op == ) mp[m]--;
             else mp[m]++;
         }
     }
 }
 int main(){
     int n;
     scanf("%d", &n);
     memset(ans, -, sizeof(ans));
     for(int i = ; i <= n; i++){
         scanf("%d%d%d", &A[i].op, &A[i].t, &A[i].m);
         A[i].id = i;
         if(A[i].op == ) ans[i] = ;
     }
     cdq(,n);
     for(int i = ; i <= n; i++){
         if(ans[i] == -) continue;
         printf("%d\n", ans[i]);
     }
     return ;
 }