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Eular Graph

TimeLimit: 1 Second   MemoryLimit: 32 Megabyte

Totalsubmit: 53   Accepted: 38  

Description

Eular graph is a graph that you can start from one vertex X and go through all the edges only once and go back to vertex X, your task is to judge whether a given graph is a Eular graph or not.

A graph will be given like this:
Give n vertexes identified from 1 to n,and a set of edges described as i,j which means that there is a edge from vertex i to vertex j. Edges are directional.

Input

Input contains serveral test cases.First line of the input file is a integer t(1<=t<=100) presents the number of test cases.Each test case contains two parts. Part one contains two integer n and m. n is the number of vertexes and m is the number of edges. Part two contains m lines with two integers(i and j)1<=i,j<=100.

Output

For each test case,print yes if the graph is a Eular graph and no if not.

Sample Input

2
5 5
1 2
2 3
3 4
4 5
5 1
4 3
1 2
2 3
3 4

Sample Output

yes
no

对于一个有向图，如果它是连通 的，而且 所有点的入度等于它的出度，则这个有向图是连通的，否则 是不连通 的。

#include<stdio.h>#include<string.h>#include<stdlib.h>int map[110][110],used[110][110],visit[110],vretex[110][110],n;void trace(int k){    int i;    for(i=1;i<=n;i++){        if(i!=k && map[k][i] == 1 && visit[i] ==0){            visit[i] = 1;            trace(i);        }    }}int main(){    int t,m,a,b,i,j;    scanf("%d",&t);    while(t--){        memset(map,0,sizeof(map));        memset(used,0,sizeof(used));        memset(vretex,0,sizeof(vretex));        memset(visit,0,sizeof(visit));        scanf("%d%d",&n,&m);        for(i=0;i<m;i++){            scanf("%d%d",&a,&b);            map[a][b] = 1;            vretex[a][0] ++ ;            vretex[b][1] ++;        }        visit[1] = 1;        trace(1);        for(i=1;i<=n;i++) if(visit[i] == 0) break;        if(i>n){            for(j=1;j<=n;j++){                if(vretex[j][0] != vretex[j][1]) break;            }            if(j>n){                printf("yes\n");                continue;            }        }        printf("no\n");    }    return 0;}