ACM: uva 10763 -&…

Foreign Exchange
 

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".

 

题意: 有n个人想当交换生, 有信息自己location和想去的location, 判断是否可以满足全部人得需求.

 

解题思路:

    
1. 简单排序即可.

 

代码:

#include <cstdio>

#include <iostream>

#include <cstring>

#include <algorithm>

using namespace std;

#define MAX 500005

struct node

{

 int A, B;

 bool operator <(const node
&x) const

 {

  if(A == x.A) return B
< x.B;

  else return A <
x.A;

 }

}a[MAX], b[MAX];

int n;

int main()

{

// freopen("input.txt", "r", stdin);

 int i;

 while(scanf("%d", &n) !=
EOF)

 {

  if(n == 0) break;

  for(i = 0; i <
n; ++i)

  {

   scanf("%d
%d", &a[i].A, &a[i].B);

   b[i].A =
a[i].B;

   b[i].B =
a[i].A;

  }

  sort(a, a+n);

  sort(b, b+n);

  bool flag = false;

  for(i = 0; i <
n; ++i)

  {

   if( a[i].A ==
b[i].A && a[i].B == b[i].B )
continue;

   else

   {

    flag
= true;

    break;

   }

  }

  if(flag)
printf("NO\n");

  else printf("YES\n");

 }

 return 0;

}