UVa 10763 - Foreign Exchange
题目链接：
原题：
Your non-profit organization (iCORE-internationalConfederation ofRevolverEnthusiasts) 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 to500000candidates!
Input
The input file contains multiple cases. Each test case will consist of a line containingn- the number of candidates(1≤n≤500000), followed bynlines representing the exchange information for each candidate. Each of these lines will contain2integers, 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 wheren=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".
题目大意：
交换生现在是很受欢迎的，现在又一个负责这个的组织，经常会收到一大批的申请表， 申请内容是从A国家到B国家的。对于一批申请表， 会有各个不同国家申请到另外各个不同的国家， 假设有任意一个申请A到B的，但是没有B到A的申请， 那么这批申请表都不能被处理。
分析与总结：
题目可以看出， 对于申请A到B的，那么一定需要有相对应的从B到A的，而且数量也必须要相对应。例如有两个A到B，但是只有一个B到A，那么也是不行的。
那么， 我们可以给出一个升序序列0,1,2,3,4,5,……n，存在数组arr中， 对于每个申请表，假设是A到B，就swap(arr[A], arr[B])。
全部处理完后， 判断这个序列是否还是和原来的排列一样。如果是的话，就符合条件。否则就不行。（因为对于每个swap（arr[A]）,arr[B]）如果有一个相对应的swap(arr[B], arr[A]), 那么相当于恢复了原排列）
/*
* UVa 10763 - Foreign Exchange
* Time : 0.132s(UVa)
* Author : D_Double
*
*/
#include<iostream>
#include<cstdio>
#include<string>
#include<algorithm>
#define MAXN 500005
using namespace std;
int arr[MAXN];
void swap(int a,int b){
int t=arr[a]; arr[a] = arr[b]; arr[b] = t;
}
void init(){
for(int i=0; i<MAXN; ++i)
arr[i] = i;
}
bool isOk(){
for(int i=0; i<MAXN; ++i)
if(arr[i]!=i) return false;
return true;
}
int main(){
int n, i, a, b;
while(scanf("%d",&n), n){
init();
for(int i=0; i<n; ++i){
scanf("%d %d", &a, &b);
swap(a, b);
}
if(isOk()) printf("YES\n");
else printf("NO\n");
}
return 0;
}
—— 生命的意义，在于赋予它意义。
原创http://blog.csdn.net/shuangde800，By D_Double (转载请标明)