POJ1386Play on Words(欧拉回路)

题目地址。

分析：

1.图必须是连通的。

2.欧拉道路或者是欧拉回路都是可以的。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define MAXN 27

int hash[MAXN], in[MAXN], out[MAXN], p[MAXN];

void Init(){
    int i;
    for(i=0; i<MAXN; i++){
        hash[i] = in[i] = out[i] = 0;
        p[i] = i;
    }
}

int find(int x){return p[x] == x ? x : (p[x] = find(p[x]));}

void Union(int u, int v){
    u = find(u); v = find(v);
    if(u != v){
        p[u] = v;
    }
}

int main(){
    int t, n, i, len, ans;
    char s[1012];

    scanf("%d", &t);

    while(t--){
        scanf("%d", &n);
        Init(n);
        for(i=0; i<n; i++){
            scanf("%s", s);
            len = strlen(s);
            int a = s[0] - 'a';
            int b = s[len-1] - 'a';
            hash[a] = 1;
            hash[b] = 1;
            out[a]++;   //不要写反了
            in[b]++;
            Union(a, b);
        }
        ans = 0;
        for(i=0; i<MAXN; i++){  //检查是否是连通图（又一个方法）
            if(hash[i] && i == find(i)) ans++;
        }

        if(ans > 1) {printf("The door cannot be opened.\n"); continue;}

        int a=0, b=0;

        for(i=0; i<MAXN; i++){
            if(hash[i] && in[i]-out[i] == 1) a++;
            else if(hash[i] && out[i]-in[i] == 1) b++;
            else if(hash[i] && out[i] != in[i]) break;
        }

        if(i<MAXN) printf("The door cannot be opened.\n");
        else if((a == 0 && b == 0) || (a == 1 && b == 1)) printf("Ordering is possible.\n");
        else printf("The door cannot be opened.\n");
    }

    return 0;
}