uva 10763 - Foreign Exchange
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Problem E
Foreign Exchange
Input: standard input
Output: standard output
Time Limit: 1 second
Your non-profit organization (iCORE -international Confederation of RevolverEnthusiasts) 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 to500000 candidates!
Input
The input file contains multiple cases. Each test case will consist of a line containingn - the number of candidates (1≤n≤500000), followed byn lines representing the exchange information for each candidate. Each of these lines will contain2 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 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".
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 100000int in[size],out[size],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[x]; ++in[y]; } f=1; for (i=1;i<=max;i++) if (in[i]!=out[i]) {f=0; break;} if (f) printf("YES\n"); else printf("NO\n"); } return 0;}
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