題目鏈接

變形課

sosoonrivergoesthemgotmoonbeginbig0 Sample Output
Yes.HintHint Harry 可以念這個咒語:"big-got-them". SourceGardon-DYGG Contest 1 RecommendJGShining   |   We have carefully selected several similar problems for you:  1175 1180 1258 1242 2553 
解題思路： DFS
用一個結構體數(shù)組a，a[i].x, a[i].y 分別用來存每個單詞的開頭和結尾， 循環(huán)查找開頭為'b'的單詞a[i].x == 'b'，
然后DFS(s)。當a[s].x == 'm'時，找到目標查詢結束，否則就繼續(xù)查找以當前點的結尾為開頭的i, 即a[s].y == a[i].x,繼續(xù)DFS(i);
代碼：
#include <cmath>#include <string>#include <cstdio>#include <sstream>#include <cstring>#include <iostream>#include <algorithm>#define LOCALusing namespace std;int n = 0;int visit[1000];int flag = 0;struct ln{    char x, y;}a[1000];void dfs(int s){    if(flag == 1) return;    //提前退出以節(jié)省時間，也可不必這也寫。        if(a[s].x == 'm')       //目標情況    {        flag = 1;        return;    }    else    {        for (int i = 0; i < n; i++)            if(a[s].y == a[i].x)    //當前點的末尾是下個點的開始            if(!visit[i])            {                visit[i] = 1;       //標記走過                dfs(i);                visit[i] = 0;       //標記還原            }    }}int main(){    string s;    while (cin>>s)    {        //因為題目要求循環(huán)輸入寫的有點亂。        if(s == "0") {            memset(visit, 0, sizeof(visit));            flag = 0;                       for(int i = 0; i < n; i++)   //循環(huán)查詢開頭為b的然后深搜            {                if(a[i].x == 'b')                    dfs(i);                if(flag == 1)                    break;            }            if(flag)                cout << "Yes." << endl;            else                cout << "No." << endl;                       n = 0;                  //記得將n置零        }        //這里為將輸入的s的開頭結尾存入a中        int k = s.length();        a[n].x = s[0];        a[n].y = s[k-1];        n++;    }    return 0;}