uva 10763
Problem EForeign 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
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.
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
Problem setter: Gilbert Lee, University of Alberta, Canada
第一眼看到题目，就有点欧拉回路的感觉，统计每个点的出度入度，0.104Ac但是回过头来看了下题目好像必须要严格的两两配对才可以，貌似不可以a-b b-c c-a，那就只有快排+二分了......数据水了
#include<string.h>
#include<stdio.h>
#define size 100000
int in,out,f,i,n,max,x,y;
int main()
{
while (scanf("%d",&n),n)
{
max=0;
memset(in,0,sizeof(in));
memset(out,0,sizeof(out));
while (n--)
{
scanf("%d%d",&x,&y);
if (max<x) max=x;
if (max<y) max=y;
++out; ++in;
}
f=1;
for (i=1;i<=max;i++)
if (in!=out) {f=0; break;}
if (f) printf("YES\n");
else printf("NO\n");
}
return 0;
}
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