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E. Little Artem and Time Machine
Little Artem has invented a time machine! He could go anywhere in time, but all his thoughts of course are with computer science. He wants to apply this time machine to a well-known data structure: multiset.
Artem wants to create a basic multiset of integers. He wants these structure to support operations of three types:
But what about time machine? Artem doesn‘t simply apply operations to the multiset one by one, he now travels to different moments of time and apply his operation there. Consider the following example.
Note that Artem dislikes exceptions so much that he assures that after each change he makes all delete operations are applied only to element that is present in the multiset. The answer to the query of the third type is computed at the moment Artem makes the corresponding query and are not affected in any way by future changes he makes.
Help Artem implement time travellers multiset.
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of Artem‘s queries.
Then follow n lines with queries descriptions. Each of them contains three integers ai, ti and xi (1 ≤ ai ≤ 3, 1 ≤ ti, xi ≤ 109) — type of the query, moment of time Artem travels to in order to execute this query and the value of the query itself, respectively. It‘s guaranteed that all moments of time are distinct and that after each operation is applied all operations of the first and second types are consistent.
For each ask operation output the number of instances of integer being queried at the given moment of time.
6
1 1 5
3 5 5
1 2 5
3 6 5
2 3 5
3 7 5
1
2
1
3
1 1 1
2 2 1
3 3 1
0
题意：
就是有一个时光机可以去不同的时刻对一些数值的数目进行修改;1是增加1,2是减1,3是询问;
思路：
一看就是个线段树,蓝而时间和数值的范围太大,时间可以进行离散化,数值的话那么就把线段树的节点变成map怒怼一发,我就是开开脑洞然后写写玩玩,蓝后一不小心怼过了;
哈哈哈,最近脑洞太大,经常乱搞来A题;我自己都受不了了;
AC代码：
/*2014300227 669E - 33 GNU C++11 Accepted 1091 ms 78840 KB*/ #include <bits/stdc++.h> using namespace std; typedef long long ll; const ll mod=1e9+7; const ll inf=1e15; const int N=1e5+6; int n; struct PO { int a,t,va,pos; }po[N]; int cmp1(PO x,PO y) { return x.t<y.t; } int cmp2(PO x,PO y) { return x.pos<y.pos; } struct Tree { int l,r; map<int,int>mp1,mp2; }; Tree tree[4*N]; void build(int node,int L,int R) { tree[node].l=L; tree[node].r=R; if(L==R)return ; int mid=(L+R)>>1; build(2*node,L,mid); build(2*node+1,mid+1,R); } void update(int node,int pos,int num,int flag) { if(tree[node].l==tree[node].r&&tree[node].l==pos) { if(flag==1) tree[node].mp1[num]++; else tree[node].mp2[num]++; return ; } int mid=(tree[node].l+tree[node].r)>>1; if(pos<=mid) { update(2*node,pos,num,flag); } else { update(2*node+1,pos,num,flag); } if(flag==1)tree[node].mp1[num]++; else tree[node].mp2[num]++; } int query(int node,int L,int R,int num,int flag) { if(L<=tree[node].l&&R>=tree[node].r) { if(flag==1)return tree[node].mp1[num]; else return tree[node].mp2[num]; } int mid=(tree[node].l+tree[node].r)>>1; if(R<=mid) { return query(2*node,L,R,num,flag); } else if(L>mid) { return query(2*node+1,L,R,num,flag); } else { return query(2*node,L,mid,num,flag)+query(2*node+1,mid+1,R,num,flag); } } int main() { scanf("%d",&n); for(int i=1;i<=n;i++) { scanf("%d%d%d",&po[i].a,&po[i].t,&po[i].va); po[i].pos=i; } sort(po+1,po+n+1,cmp1); for(int i=1;i<=n;i++) { po[i].t=i; } sort(po+1,po+n+1,cmp2); build(1,1,n); for(int i=1;i<=n;i++) { if(po[i].a==1) { update(1,po[i].t,po[i].va,1); } else if(po[i].a==2) { update(1,po[i].t,po[i].va,2); } else { //cout<<"#"<<endl; printf("%d\n",query(1,1,po[i].t,po[i].va,1)-query(1,1,po[i].t,po[i].va,2)); } } return 0; }
codeforces 669E E. Little Artem and Time Machine(节点为map型的线段树)
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原文地址：http://www.cnblogs.com/zhangchengc919/p/5433262.html
