HDU 4022 Bombing c++解法

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 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.

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.

写一下本人的解法，一开始想到用结构体去写，写完一个O(N)的，oj上超时。后面用map写O(logn)的，也是超时。最后把cin改成scanf才能ac。说明这道题的测试样例点很多。而scanf读入速度约为cin的10倍，这题时间要求挺严格的。

下面粘贴代码：

#include <map>
#include <iostream>
#include <set>
#include <cstdio>
using namespace std;int main() {int n, m;while (cin >> n >> m) {map<long long int, multiset<long long int > > mapx;map<long long int, multiset<long long int > > mapy;if (n == 0 && m == 0)break;long long int x, y;for (int i = 0; i < n; i++) {scanf("%lld %lld", &x, &y);mapx[y].insert(x);mapy[x].insert(y);}multiset<long long int >::iterator it;long long int choose, line;for (int i = 0; i < m; i++) {cin >> choose >> line;if (choose == 1) {cout << mapx[line].size() << endl;//用y找有几个x，输出对应数量for (it = mapx[line].begin(); it != mapx[line].end(); it++) {//输出完之后，把另一个map中的对应点删掉if (mapy[*it].empty())//不判断empty可能会报错，我的编译器会报错。continue;elsemapy[*it].erase(mapy[*it].find(line));}mapx[line].clear();//删除y=line时所有的x}if (choose == 0) {//同理用x找有几个y时cout << mapy[line].size() << endl;for (it = mapy[line].begin(); it != mapy[line].end(); it++) {if (mapx[*it].empty())continue;elsemapx[*it].erase(mapx[*it].find(line));}mapy[line].clear();}}cout << endl;}return 0;
}

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