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
这题目用哈希做刚刚好， 注意优化的是凡是匹配的点以后就可以不用匹配了，否则很容易就超时
#include <stdio.h>
#include <string.h>
#include <math.h>
struct num
{
char s1,s2;
int next,status,pos;
}a;
int b;
int prim=1000001;
struct Num
{
char s1,s2;
}c;
int status;
int main()
{
int i,j,n,m,s,t,tag,s1,s2,yushu,x,l,key;
while(scanf("%d",&n)!=EOF)
{
if(n==0)
{
break;
}
memset(b,-1,sizeof(b));
for(i=0,tag=0;i<=n-1;i++)
{
scanf("%s %s",c.s1,c.s2);
l=strlen(c.s1);
for(j=0,s1=0;j<=l-1;j++)
{
s1=(s1*10%prim+c.s1-'0')%prim;
}
l=strlen(c.s2);
for(j=0,s2=0;j<=l-1;j++)
{
s2=(s2*10%prim+c.s2-'0')%prim;
}
yushu=(s1*10%prim+s2%prim)%prim;
strcpy(a.s1,c.s1);
strcpy(a.s2,c.s2);
a.next=b; a.pos=i;
a.status=0;b=tag;
tag++;
}
memset(status,0,sizeof(status));
for(i=0;i<=n-1;i++)
{
if(!status)
{
l=strlen(c.s2);
for(j=0,s2=0;j<=l-1;j++)
{
s2=(s2*10%prim+c.s2-'0')%prim;
}
l=strlen(c.s1);
for(j=0,s1=0;j<=l-1;j++)
{
s1=(s1*10%prim+c.s1-'0')%prim;
}
yushu=(s2*10%prim+s1%prim)%prim;
for(x=b,key=0;x!=-1; x=a.next)
{
if(!a.status&&!status.pos])
{
if(strcmp(a.s1,c.s2)==0&&strcmp(a.s2,c.s1)==0)
{
key=1;
a.status=1;
status=1;
status.pos]=1;
break;
}
}
}
if(!key)
{
break;
}
}
}
if(i==n)
{
printf("YES\n");
}else
{
printf("NO\n");
}
}
return 0;
}
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