UVa 208

这道题代码很容易写出来，难点在于很容易超时（特别是起点端图特别稠密的情况下，终点与起点没有连接），解决方法是先从终点找连接点。

#include <cstdio>
#include <cstring>
int G[25][25];
int vis[25];
int join[22];
int e;
int r=0;
void find_join(int u){
    join[u]=1;
    for(int v=0;v<25;v++){
        if(!join[v] && G[u][v])
            find_join(v);
    }
}
int dfs(int s,int k,int *a){
    a[k]=s;
    if(s==e){
        for(int i=0;i<k;i++){
            printf("%d ",a[i]);
        }
        r++;
        if(k>=0)printf("%d",a[k]);
        printf("
");
        return 1;
    }
    int ok=0;
    for(int i=0;i<25;i++){
        if(G[s][i]&&!vis[i]&&join[i]){
            vis[i]=1;
            if(dfs(i,k+1,a))ok=1;
            vis[i]=0;
        }
    }
    if(ok)return 1;
    else return 0;
}
int main(){
    int cnt = 0;
    while(scanf("%d",&e)!=EOF){
        memset(G,0,sizeof(G));
        memset(join,0,sizeof(join));
        memset(vis,0,sizeof(vis));
        int a,b;
        while(scanf("%d%d",&a,&b)==2&&(a||b)){
            G[a][b]=1;
            G[b][a]=1;
        }

        int t[25];
        vis[1]=1;
        printf("CASE %d:
",++cnt);
        r=0;
        find_join(e);
        dfs(1,0,t);
        printf("There are %d routes from the firestation to streetcorner %d.
",r,e);
    }
    return 0;
}