uva 11396 二分图判定

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<vector>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<map>
#include<stack>

using namespace std;

#define LL long long
#define UINT unsigned int
#define MAX_INT 0x7fffffff
#define cint const int

#define MAXN 333
vector<int> g[MAXN];
int n, col[MAXN];

bool dfs(int u, int color){
    col[u] = color;
    for(int i=0; i<g[u].size(); i++){
        int v = g[u][i];
        if(col[v]==col[u] || !col[v]&&!dfs(v, 3-color))
            return false;
    }
    return true;
}

int main(){
//    freopen("C:\\Users\\Administrator\\Desktop\\in.txt","r",stdin);
    while(scanf(" %d", &n)==1 && n){
        int i, u, v;
        for(i=1; i<=n; i++) g[i].clear();
        while(scanf(" %d %d", &u, &v)==2){
            if(!v) break;
            g[u].push_back(v);
            g[v].push_back(u);
        }
        fill_n(col+1, n, 0);
        for(i=1; i<=n; i++) if(!col[i])
            if(!dfs(i, 1)) break;
        if(i<=n) printf("NO\n");
        else printf("YES\n");
    }
    return 0;
}

Problem B
Claw Decomposition

Input: Standard Input

Output: Standard Output

 

A claw is defined as a pointed curved nail on the end of each toe in birds, some reptiles, and some mammals. However, if you are a graph theory enthusiast, you may understand the following special class of graph as shown in the following figure by the word claw.

If you are more concerned about graph theory terminology, you may want to define claw as K1,3.

 

Let’s leave the definition for the moment & come to the problem. You are given a simple undirected graph in which every vertex has degree 3. You are to figure out whether the graph can be decomposed into claws or not.

 

Just for the sake of clarity, a decomposition of a graph is a list of subgraphs such that each edge appears in exactly one subgraph in the list.

 

Input

 

There will be several cases in the input file. Each case starts with the number of vertices in the graph, V (4<=V<=300). This is followed by a list of edges. Every line in the list has two integers, a & b, the endpoints of an edge (1<=a,b<=V). The edge list ends with a line with a pair of 0. The end of input is denoted by a case with V=0. This case should not be processed.

 

Output

 

For every case in the input, print YES if the graph can be decomposed into claws & NO otherwise.

 

Sample Input                                                  Output for Sample Input

4

1 2

1 3

1 4

2 3

2 4

3 4

0 0

6

1 2

1 3

1 6

2 3

2 5

3 4

4 5

4 6

5 6

0 0

0

NO

NO

 

 


Problemsetter: Mohammad Mahmudur Rahman

Special Thanks to: Manzurur Rahman Khan

 

参考:http://blog.csdn.net/wiking__acm/article/details/8739684

。。。。题意都不大懂,先挖个坑。。。虽然A了。。。
posted @ 2013-10-20 01:31  Ramanujan  阅读(256)  评论(0编辑  收藏  举报