题目链接：https://vjudge.net/problem/HDU-4022
题目描述：
InputMultiple 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 10 9, 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.
OutputFor 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
用两个map来模拟，一个x到y的映射，一个y到x的映射，关于怎么解决删除重复元素的问题，可以在每次删除一个映射的元素的时候，删除掉对应的另外一个映射的元素
#pragma G++ optimize(2) #include<set> #include<map> #include<stack> #include<queue> #include<cmath> #include<stdio.h> #include<cctype> #include<string> #include<vector> #include<climits> #include<cstring> #include<cstdlib> #include<iostream> #include<algorithm> #define endl ‘\n‘ #define max(a, b) (a > b ? a : b) #define min(a, b) (a < b ? a : b) #define mst(a) memset(a, 0, sizeof(a)) #define _test printf("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n") using namespace std; typedef long long ll; typedef pair<int, int> P; const double eps = 1e-7; const int INF = 0x3f3f3f3f; const ll ll_INF = 0x3f3f3f3f3f3f3f; const int maxn = 1e3 + 10; map<int, multiset<int> > xbase; //x到y的映射 map<int, multiset<int> > ybase; //y到x的映射 typedef map<int, multiset<int> > ite; int booms(ite &x, ite &y, int pos) { for (multiset<int>::iterator it = x[pos].begin(); it != x[pos].end(); ++it) y[*it].erase(pos); //删除被炸弹炸掉的元素 int res = x[pos].size(); //炸弹炸掉的基地数量 x[pos].clear(); //模拟炸弹清空对应的行(列) return res; } int main(void) { //std::ios::sync_with_stdio(false); int n, m; while(~scanf("%d%d", &n, &m) && (n||m)) { for (int i = 0, x, y; i<n; ++i) { scanf("%d%d", &x, &y); xbase[x].insert(y); ybase[y].insert(x); } for (int i = 0, c, d; i<m; ++i) { scanf("%d%d", &c, &d); if (!c) printf("%d\n", booms(xbase, ybase, d)); else printf("%d\n", booms(ybase, xbase, d)); } xbase.clear(); ybase.clear(); putchar(endl); } return 0; }
原文：https://www.cnblogs.com/shuitiangong/p/12210461.html