UVA - 10763
Description 
Problem E 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! InputThe input file contains multiple cases. Each test case will consist of a line containing n - the number of candidates (1≤n≤500000), followed by n lines 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.
OutputFor 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
Problem setter: Gilbert Lee, University of Alberta, Canada Source Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: Problem Solving Paradigms :: Greedy :: Involving
Sorting (Or The Input Is Already Sorted)
Root :: AOAPC I: Beginning Algorithm Contests (Rujia Liu) :: Volume 4. Algorithm Design Root :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: Problem Solving Paradigms ::Greedy - Standard Root :: Competitive Programming: Increasing the Lower Bound of Programming Contests (Steven & Felix Halim) :: Chapter 3. Problem Solving Paradigms :: Greedy Root :: AOAPC II: Beginning Algorithm Contests (Second Edition) (Rujia Liu) :: Chapter 5. C++ and STL :: Exercises Root :: AOAPC I: Beginning Algorithm Contests -- Training Guide (Rujia Liu) :: Chapter 1. Algorithm Design :: Designing Efficient Algorithms :: Exercises: Beginner |
借鉴Staginner的想法
AC代码：
#include <cstdio>
#include <cstring>
using namespace std;
int G[1000][1000], n;
int check()
{
int i,j;
for(i=0;i<1000;i++)
for(j=0;j<1000;j++)
if(G[i][j]!=0)
return 0;
return 1;
}
int main()
{
int i, j, u, v;
while(1)
{
scanf("%d",&n);
if(n==0)
break;
memset(G, 0, sizeof(G));
for(i=0;i<n;i++)
{
scanf("%d%d",&u,&v);
G[u][v]++;
G[v][u]--;
}
if(check())
printf("YES\n");
else
printf("NO\n");
}
return 0;
}UVA - 10763 - Foreign Exchange
原文地址：http://blog.csdn.net/u014355480/article/details/41829851
