zoj 2475 Benny's Compiler

1、http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2475

2、题目大意：

题目给了一个图，包含各边的关系，现在 要看一个点是否在一个环内，若在输出no,不在输出yes,

dfs(),如果搜到重复的点，说明在环内，否则不在

3、题目：

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

4、AC代码：

#include<stdio.h>
#include<string.h>
int map[105][105];
int vis[105];
int flag,n;
void dfs(int a)
{
    for(int i=1;i<=n;i++)
    {
        if(vis[i]==1 && map[a][i]==1)
        {
            flag=0;
            return ;
        }
        if(map[a][i]==1)
        {
            vis[i]=1;
            dfs(i);
            vis[i]=0;
        }
    }
}
int main()
{
    int a,b;
    while(scanf("%d",&n)!=EOF)
    {
        if(n==-1)
        break;
        memset(map,0,sizeof(map));
        memset(vis,0,sizeof(vis));
        for(int i=0; i<n; i++)
        {
            scanf("%d%d",&a,&b);
            if(a!=b)
               map[a][b]=1;
        }
        scanf("%d",&a);
        vis[a]=1;
        flag=1;
        dfs(a);
        if(flag)
        printf("Yes\n");
        else
        printf("No\n");
    }
    return 0;
}