ZOJ 2475 Benny's Compiler 有向图判读点是否存在环内

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Benny's Compiler
Time Limit: 2 Seconds      Memory Limit: 65536 KB
These days Benny has designed a new compiler for C programming language. His compilation system provides a compiler driver that invokes the language preprocessor, compiler, assembler
and linker. C source file (with .C suffix) is translated to relocatable object module first, and then all modules are linked together to generate an executable object file.
The translator (preprocessor, compiler and assembler) works perfectly and can generate well optimized assembler code from C source file. But the linker has a serious bug -- it cannot
resolve global symbols when there are circular references. To be more specific, if file 1 references variables defined in file 2, file 2 references variables defined in file 3, ... file n-1 references variables defined in file n and file n references variables
defined in file 1, then Benny's linker walks out because it doesn't know which file should be processed first.
Your job is to determine whether a source file can be compiled successfully by Benny's compiler.

Input
There are multiple test cases! In each test case, the first line contains one integer N, and then N lines follow. In each of these lines there are two integers Ai and Bi, meaning that
file Ai references variables defined in file Bi (1 <= i <= N). The last line of the case contains one integer E, which is the file we want to compile.
A negative N denotes the end of input. Else you can assume 0 < N, Ai, Bi, E <= 100.

Output
There is just one line of output for each test case. If file E can be compiled successfully output "Yes", else output "No".

Sample Input
4

1 2

2 3

3 1

3 4

1
4

1 2

2 3

3 1

3 4

4
-1

Sample Output
No

Yes

Author: ZHENG, Lu
Source: Zhejiang Provincial Programming Contest 2005

给你n个点，然后给你点与点的连接关系，构成一个有向图，让你判读一个点是否在一个环内。若在，则输出No，否则输出Yes。
用dfs深搜，如果搜到已经访问过的点，则说明该点在环内，否则不再环内。

#include<stdio.h>
#include<string.h>
#define M 107
int g[M][M],vis[M],flag,n;
void dfs(int u)
{
for(int i=1;i<=n;i++)
{
if(vis[i]&&g[u][i]){flag=0;return;}
if(g[u][i])
{
vis[i]=1;
dfs(i);
vis[i]=0;
}
}
}
int main()
{
while(scanf("%d",&n)!=EOF&&n!=-1)
{

memset(vis,0,sizeof(vis));
memset(g,0,sizeof(g));
int u,v;
flag=1;
for(int i=1;i<=n;i++)
{
scanf("%d%d",&u,&v);
if(u!=v)//如果不加，会WA,已验证
g[u][v]=1;
}
scanf("%d",&u);
vis[u]=1;
dfs(u);
if(flag)printf("Yes\n");
else printf("No\n");
}
return 0;
}