Codeforces 669E Little Artem and Time Machine (离散化树状数组)

题意

给出一个multiset，可以对他进行三种操作:
1.在t时间增加一个x
2.在t时间删除一个x
3.查询t时间x的个数并输出

思路

题意看了很久才懂= =，看懂之后决定朴素BIT来一发，因为有时间和x两个维度，且都是1e9，选择其中任意一个离散化一下就可以用map来维护sum了，然后过了。时间390ms，内存50000+。。。

代码

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
#define LL long long
#define lowbit(x) ((x)&(-x))
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1|1
#define MP(a, b) make_pair(a, b)
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int maxn = 1e5 + 10;
const double eps = 1e-8;
const double PI = acos(-1.0);
typedef pair<int, int> pii;
map<int, int> sum[100010], vis;

void update(int time, int x, int val)
{
    while (time < MOD)
    {
        sum[x][time] += val;
        time += lowbit(time);
    }
}

int query(int time, int x)
{
    int ans = 0;
    while (time > 0)
    {
        ans += sum[x][time];
        time -= lowbit(time);
    }
    return ans;
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int n;
    scanf("%d", &n);
    int id = 0;
    vis.clear();
    for (int i = 0; i < n; i++)
    {
        int op, t ,x;
        scanf("%d%d%d", &op, &t, &x);
        if (op == 1 && !vis[x]) vis[x] = ++id;
        if (op == 1)
            update(t, vis[x], 1);
        if (op == 2)
            update(t, vis[x], -1);
        if (op == 3)
            printf("%d\n", query(t, vis[x]));
    }
    return 0;
}