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
題意：有n個交換生。然後接下去表示每個學生能從A校到B校。。只有當一個學生從A校到B校。另一個學生從B校到A校。才可以進行交換。也就是說兩兩配對。。最後求出是否能全部進行交換。
思路：這題n有50W。如果直接開二維數組。肯定不行的。於是用vector存鄰接表。然後就是判斷能不能兩兩配對的。詳細看代碼。
#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 --;
}//如果有一個配對的。就把配對的那個從關系中刪除。
else {
map[a].push_back(b);
judge ++;
}//如果沒有配對的。這當前這個關系保存起來。
}
if (judge)//如果是兩兩配對。最後judge是為0的
printf("NO\n");
else
printf("YES\n");
}
return 0;
}
