UVA - 10763 Foreign Exchange(vector的使用)

Problem E
Foreign Exchange
Input: standard input
Output: standard output
Time Limit: 1 second

Your non-profit organization (iCORE - international Confederation of Revolver Enthusiasts) coordinates a very successful foreign student exchange program. Over the last few years, demand has sky-rocketed and now you need assistance with your task.

The program your organization runs works as follows: All candidates are asked for their original location and the location they would like to go to. The program works out only if every student has a suitable exchange partner. In other words, if a student wants to go from A to B, there must be another student who wants to go from B to A. This was an easy task when there were only about 50 candidates, however now there are up to 500000 candidates!

Input

The input file contains multiple cases. Each test case will consist of a line containing n - the number of candidates (1≤n≤500000), followed by nlines representing the exchange information for each candidate. Each of these lines will contain 2 integers, separated by a single space, representing the candidate's original location and the candidate's target location respectively. Locations will be represented by nonnegative integer numbers. You may assume that no candidate will have his or her original location being the same as his or her target location as this would fall into the domestic exchange program. The input is terminated by a case where n = 0; this case should not be processed.

 

Output

For each test case, print "YES" on a single line if there is a way for the exchange program to work out, otherwise print "NO".

 

Sample Input                               Output for Sample Input

10 1 2 2 1 3 4 4 3 100 200 200 100 57 2 2 57 1 2 2 1 10 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 0

YES NO

题意：有n个交换生。然后接下去表示每个学生能从A校到B校。。只有当一个学生从A校到B校。另一个学生从B校到A校。才可以进行交换。也就是说两两配对。。最后求出是否能全部进行交换。

思路：这题n有50W。如果直接开二维数组。肯定不行的。于是用vector存邻接表。然后就是判断能不能两两配对的。详细看代码。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
using namespace std;

int n, i;
int a, b;
int judge;
vector<int> map[500005];

int main() {
	while (~scanf("%d", &n) && n) {
		judge = 0;
		memset(map, 0, sizeof(map));
		for (i = 0; i < n; i ++) {
			scanf("%d%d", &a, &b);
			if (find(map[b].begin(), map[b].end(), a) != map[b].end()) {
				map[b].erase(find(map[b].begin(), map[b].end(), a));
				judge --;
			}//如果有一个配对的。就把配对的那个从关系中删除。
			else {
				map[a].push_back(b);
				judge ++;
			}//如果没有配对的。这当前这个关系保存起来。
		}
		if (judge)//如果是两两配对。最后judge是为0的
			printf("NO\n");
		else
			printf("YES\n");
	}
	return 0;
}