Day3—A

It’s a cruel war which killed millions of people and ruined series of cities. In order to stop it, let’s bomb the opponent’s base.
It seems not to be a hard work in circumstances of street battles, however, you’ll be encountered a much more difficult instance: recounting exploits of the military. In the bombing action, the commander will dispatch a group of bombers with weapons having the huge destructive power to destroy all the targets in a line. Thanks to the outstanding work of our spy, the positions of all opponents’ bases had been detected and marked on the map, consequently, the bombing plan will be sent to you.
Specifically, the map is expressed as a 2D-plane with some positions of enemy’s bases marked on. The bombers are dispatched orderly and each of them will bomb a vertical or horizontal line on the map. Then your commanded wants you to report that how many bases will be destroyed by each bomber. Notice that a ruined base will not be taken into account when calculating the exploits of later bombers.

Input

Multiple test cases and each test cases starts with two non-negative integer N (N<=100,000) and M (M<=100,000) denoting the number of target bases and the number of scheduled bombers respectively. In the following N line, there is a pair of integers x and y separated by single space indicating the coordinate of position of each opponent’s base. The following M lines describe the bombers, each of them contains two integers c and d where c is 0 or 1 and d is an integer with absolute value no more than 109, if c = 0, then this bomber will bomb the line x = d, otherwise y = d. The input will end when N = M = 0 and the number of test cases is no more than 50.

Output

For each test case, output M lines, the ith line contains a single integer denoting the number of bases that were destroyed by the corresponding bomber in the input. Output a blank line after each test case.

Sample Input

3 2
1 2
1 3
2 3
0 1
1 3
0 0
Sample Output

2
1
题意：N个坐标，有M次操作，每次输入d,x，当d=0时，删除x轴上坐标为x的所有坐标并输出个数，d=1时，删除y轴上坐标为x的所有坐标输出个数。

思路：用map+multiset，一个存x轴上的每个坐标对应的y坐标，一个存y轴。

代码如下：

#include<bits/stdc++.h>
using namespace std;
#define ma map<int,multiset<int> >
#define se multiset<int>
ma a,b;
void my_solve(se &r1,ma &r2,int d)
{
    int cnt=0;
    for(set<int>::iterator it=r1.begin();it!=r1.end();it++)
    {
        r2[*it].erase(d);
        cnt++;
    }
    r1.clear();
    cout<<cnt<<endl;
}
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m)&&(n!=0||m!=0))
    {
        int x,y;
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d",&x,&y);
            a[x].insert(y);
            b[y].insert(x);
        }
        for(int i=1;i<=m;i++)
        {
            int c,d,cnt=0;
            scanf("%d%d",&c,&d);
            if(c==0)
                my_solve(a[d],b,d);
            else
                my_solve(b[d],a,d);
        }
        printf("\n");
        a.clear();
        b.clear();
    }
    return 0;
}