1320. Graph Decomposition

1320. Graph Decomposition

Time limit: 0.5 second Memory limit: 64 MB

There is a simple graph with an even number of edges. You are to define if it is possible to present it by the set of pairs of adjacent edges (having a common vertex).

Input

contains a sequence of the numbers pairs. Each pair denotes vertices identifiers of one edge. All the identifiers are integers from 1 to 1000. You may assume that there are no loops and multiple edges in the graph defined by the input data.

Output

“1” (without quotation marks), if the decomposition is possible and “0” otherwise.

Samples

input	output
1 2 2 3 3 1 1 10	1
1 2 2 3 3 1 4 10	0

Problem Author: Idea: Alexander Petrov, prepared by Alexander Petrov, Leonid Volkov Problem Source: VIII Collegiate Students Urals Programming Contest. Yekaterinburg, March 11-16, 2004

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并查集&&计算相应根节点的度数和。

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#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<cstdio>
using namespace std;
int fa[10001];
int de[10001];
int a,b,i,j,k;
void make()
{
    for(i=1;i<=10001;i++)
     fa[i]=i;
}
int find(int x)//并查集
 {
     if(fa[x]!=x)
      fa[x]=find(fa[x]);
     if(fa[x]==x)
      return fa[x];
}
void uon(int x,int y)
 {
     x=find(fa[x]);
     y=find(fa[y]);
     if(x!=y)
      {
          fa[y]=x;
          de[x]+=de[y];//计算相应根节点的度数
          de[y]=0;
      }
 }
int main()
{
    make();
    memset(de,0,sizeof(de));
    while(scanf("%d %d",&a,&b)!=EOF)
     {
         uon(a,b);
         de[find(b)]++;
     }
     for(i=1;i<10001;i++)
      if(de[i]%2)//不为偶数不满足条件
       {
           cout<<'0'<<endl;
           return 0;
       }
     cout<<'1'<<endl;
     return 0;

}