标签:chmod arc hint ane else 单词 tle bottom cst
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 23536 Accepted Submission(s): 8532
#include<iostream> #include<cstring> #include<queue> #include<cstdio> #include<algorithm> using namespace std; char s[10010][1010]; int n,visit[10010]; int dfs(char c) { if(c==‘m‘) return 1;//结束标志为搜的字母等于m for(int i=1;i<=n;i++) { if(!visit[i]&&c==s[i][0]) { visit[i] = 1; if(dfs(s[i][strlen(s[i])-1])) return 1; } } } int main() { while(scanf("%s",s[1])!=EOF) { n = 1; while(scanf("%s",s[++n])!=EOF&&s[n][0]!=‘0‘);//这样输入并判断输入是否为零,简单清晰明了 memset(visit,0,sizeof(visit)); int i; for(i=1;i<=n;i++) { if(s[i][0]==‘b‘) { visit[i] = 1; if(dfs(s[i][strlen(s[i])-1])) break;//直接搜一个字母就可以了 } } if(i>n) printf("No.\n"); else printf("Yes.\n"); } return 0; }
标签:chmod arc hint ane else 单词 tle bottom cst
原文地址:http://www.cnblogs.com/l609929321/p/6596990.html