ACM/ICPC 之 DFS求解欧拉通路路径（POJ2337）

　　判断是欧拉通路后，DFS简单剪枝求解字典序最小的欧拉通路路径

//Time:16Ms     Memory:228K
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;

#define MAX 1005
#define MAXS 24 //姓名
#define MAXN 26 //字母

struct Edge{
    char name[MAXS];
    int a,b;
    friend bool operator < (Edge &e1, Edge &e2)
    { return strcmp(e1.name, e2.name) < 0; }
}e[MAX];

int n;
int in[MAXN], out[MAXN];    //入度与出度
int order[MAX];     //顺序路径
bool v[MAX];

bool dfs(int x, int rk)
{
    if(rk == n) return true;
    for(int i = 0; i < n; i++)
    {
        if(!v[i] && x == e[i].a){
            v[i] = true;
            order[rk] = i;
            if(dfs(e[i].b, rk+1)) return true;
            v[i] = false;
        }
    }
    return false;
}

int main()
{
    //freopen("in.txt","r",stdin);

    int T;
    scanf("%d",&T);
    while(T--){
        memset(e,0,sizeof(e));
        memset(in,0,sizeof(in));
        memset(out,0,sizeof(out));
        memset(order,-1,sizeof(order));
        memset(v,false,sizeof(v));
        scanf("%d",&n);
        //构图
        int st = 26;    //st:起点
        for(int i = 0; i < n; i++)
        {
            scanf("%s",e[i].name);
            e[i].a = e[i].name[0] - ‘a‘;
            e[i].b = e[i].name[strlen(e[i].name) - 1] - ‘a‘;
            out[e[i].a]++;
            in[e[i].b]++;
            if(e[i].a < st) st = e[i].a;
        }
        sort(e,e+n);
        //欧拉路判定
        int odd = 0;    //奇度结点个数
        bool flag = true;
        for(int i = 0 ; i < MAXN; i++)
        {
            if(in[i] != out[i])
            {
                odd++;
                if(out[i] - in[i] == 1)
                    st = i;
                else if(out[i] - in[i] != -1)
                {
                    flag = false;
                    break;
                }
            }
        }
        if(flag && (odd ==2 || odd == 0) && dfs(st,0))  //满足欧拉通路(排除非连通)+完整路径(判断非连通)
        {
            for(int i = 0; i < n - 1; i++)
                printf("%s.", e[order[i]].name);
            printf("%s\n", e[order[n-1]].name);
        }
        else printf("***\n");

    }

    return 0;
}