UVa11396 - Claw Decomposition

最新推荐文章于 2019-09-07 13:24:49 发布

原创最新推荐文章于 2019-09-07 13:24:49 发布 · 753 阅读

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本文探讨了一个特定的图论问题——爪分解。对于每个顶点度数为3的简单无向图，判断其是否能被分解成一系列的爪子（即K1,3）。输入包括多个图实例，每个实例包含顶点数及边的信息。输出则针对每个实例给出YES或NO的答案。

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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 K_1,3.

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

1 2

1 3

1 4

2 3

2 4

3 4

0 0

1 2

1 3

1 6

2 3

2 5

3 4

4 5

4 6

5 6

0 0

Problemsetter: Mohammad Mahmudur Rahman

Special Thanks to: Manzurur Rahman Khan

判断二部图用的DFS 其实BFS更佳，因为不容易溢出

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
using namespace std;

const int maxn = 300+10;
int head[maxn];
int color[maxn];
int n,cnt;
struct Edge{
    int nxt,to;
}edge[maxn*maxn];

void addedge(int u,int v){
    cnt++;
    edge[cnt].nxt = head[u];
    edge[cnt].to = v;
    head[u] = cnt;
}

bool dfs(int x){
    for(int i = head[x]; i != -1;i = edge[i].nxt){
        int v = edge[i].to;
        if(color[v] == color[x]) return 0;
        if(!color[v]){
            color[v] = 3-color[x];
            if(!dfs(v)) return false;
        }
    }
    return true;
}

void init(){
    memset(head,-1,sizeof head);
    memset(color,0,sizeof color);
    cnt = 0;
}

int main(){
    while(cin >> n&&n){
        init();
        int a,b;
        while(cin >> a >> b && a+b){
                addedge(a,b);
                addedge(b,a);
        }
        color[1] = 1;
        if(dfs(1)){
            cout<<"YES"<<endl;
        }else{
            cout<<"NO"<<endl;
        }
    }
    return 0;
}