poj 2337 Catenyms 欧拉回路+dfs 解题报告

Catenyms

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 11494 Accepted: 2982

Description

A catenym is a pair of words separated by a period such that the last letter of the first word is the same as the last letter of the second. For example, the following are catenyms: 

dog.gophergopher.ratrat.tigeraloha.alohaarachnid.dog

A compound catenym is a sequence of three or more words separated by periods such that each adjacent pair of words forms a catenym. For example, 

aloha.aloha.arachnid.dog.gopher.rat.tiger 

Given a dictionary of lower case words, you are to find a compound catenym that contains each of the words exactly once.

Input

The first line of standard input contains t, the number of test cases. Each test case begins with 3 <= n <= 1000 - the number of words in the dictionary. n distinct dictionary words follow; each word is a string of between 1 and 20 lowercase letters on a line by itself.

Output

For each test case, output a line giving the lexicographically least compound catenym that contains each dictionary word exactly once. Output "***" if there is no solution.

Sample Input

26alohaarachniddoggopherrattiger3oakmapleelm

Sample Output

aloha.arachnid.dog.gopher.rat.tiger***

题意：求欧拉路径，单词首尾相连，代码参考这位大神

用例：
1
6
aloha
arachnid
dog
gopher
rat
riger
这个用例说明为什么要倒着记录路径，当找r时，rat是最小的，然后去找t，找不到，所以rat最先入结果集，所以rat为最后一个。

代码：

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <queue>using namespace std;struct Edge{    int to,next;    int index;    bool flag;}edge[2010];int head[30],tot;void addedge(int u,int v,int index){    edge[tot].to = v;    edge[tot].next = head[u];    edge[tot].index = index;    edge[tot].flag = false;    head[u] = tot++;}string str[1010];int in[30],out[30];int cnt;int ans[1010];void dfs(int u){    for(int i = head[u]; i != -1; i = edge[i].next)        if(!edge[i].flag)        {            edge[i].flag = true;            dfs(edge[i].to);            ans[cnt++] = edge[i].index;        }}int main(){    int T, n;    scanf("%d", &T);    while(T--)    {        scanf("%d", &n);        for(int i = 0; i < n; i++)            cin >> str[i];        sort(str, str + n);        tot = 0;        memset(head, -1, sizeof(head));        memset(in, 0, sizeof(in));        memset(out, 0, sizeof(out));        int start = 100;        for(int i = n - 1; i >= 0; i--)//字典序大的先加入        {            int u = str[i][0] - 'a';            int v = str[i][str[i].length() - 1] - 'a';            addedge(u, v, i);            out[u]++;            in[v]++;            if(u < start)start = u;            if(v < start)start = v;        }        int cc1 = 0, cc2 = 0;        for(int i = 0;i < 26;i++)        {            if(out[i] - in[i] == 1)            {                cc1++;                start = i;//如果有一个出度比入度大1的点，就从这个点出发，否则从最小的点出发            }            else if(out[i] - in[i] == -1)                cc2++;            else if(out[i] - in[i] != 0)                cc1 = 3;        }        if(! ( (cc1 == 0 && cc2 == 0) || (cc1 == 1 && cc2 == 1) ))        {            printf("***\n");            continue;        }        cnt = 0;        dfs(start);        if(cnt != n)//判断是否连通        {            printf("***\n");            continue;        }        for(int i = cnt-1; i >= 0;i--)        {            cout<<str[ans[i]];            if(i > 0)printf(".");            else printf("\n");        }    }    return 0;}