听听 艾薇儿 girlfriend

zoj 2475 Benny's Compiler

ZOJ Problem Set - 2475
Benny's Compiler

Time Limit: 1 Second      Memory Limit: 32768 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
Submit    Status

//1859524 2009-05-08 09:52:20 Time Limit Exceeded  2475 C++ 1001 0 Wpl
//1859528 2009-05-08 09:56:35 Memory Limit Exceeded  2475 C++ 0 32769 Wpl 
//1859534 2009-05-08 10:00:46 Wrong Answer  2475 C++ 0 224 Wpl 
//1859641 2009-05-08 11:10:30 Accepted  2475 C++ 0 224 Wpl 
#include <iostream>
#include 
<queue>
#define MAX 102
using namespace std;
int edges[MAX][MAX];
int n,start;
bool mark,used[MAX];
void Init()
{
    
int i,a,b,j;
    
for(i=1;i<MAX;i++)
        
for(j=1;j<MAX;j++)
            edges[i][j]
=-1;
    
for(i=0;i<n;i++)
    {
        scanf(
"%d%d",&a,&b);
        
if(a!=b)
            edges[a][b]
=1;
    }
}
void DFS(int start)
{
    
int i;
    
if(used[start])
    {
        mark
=true;
        
return;
    }
    used[start]
=true;
    
for(i=1;i<MAX;i++)
    {
        
if(edges[start][i]==1)
        {
            DFS(i);
        }
    }
    used[start]
=false;
}
int main()
{
    
int i;
    
while(scanf("%d",&n)!=EOF&&n>=0)
    {
        Init();
        scanf(
"%d",&start);
        mark
=false;
        
for(i=1;i<MAX;i++)
            used[i]
=false;
        DFS(start);
        
if(!mark)
            printf(
"Yes\n");
        
else
            printf(
"No\n");
    }
    
return 0;
}

 
 
//1859782 2009-05-08 13:05:57 Accepted  2475 C++ 0 184 Wpl 
//邻接表
#include <iostream>
#include 
<vector>
#define MAX 102
using namespace std;
vector
<int>G[MAX];
vector
<int>::iterator p;
int n,start;
bool mark,used[MAX];
void Init()
{
    
int i;
    
int a,b;
    
for(i=1;i<MAX;i++)
    {
        G[i].clear();
        used[i]
=false;
    }
    
for(i=1;i<=n;i++)
    {
        scanf(
"%d%d",&a,&b);
        
if(a!=b)          //a a 或者b b这种不算是环的
            G[a].push_back(b);
    }
}
void DFS(int start)
{
    
if(used[start])
    {
        mark
=true;
        
return;
    }
    used[start]
=true;
    
/*if(G[start].empty())
        return ;
    for(p=G[start].begin();p!=G[start].end();p++)
        DFS(*p);
*/
    
int i;
    
for(i=0;i<G[start].size();i++)
        DFS(G[start][i]);
    used[start]
=false;
}
int main()
{
    
while(scanf("%d",&n)!=EOF&&n>=0)
    {
        Init();
        scanf(
"%d",&start);
        mark
=false;
        DFS(start);
        
if(!mark)
            printf(
"Yes\n");
        
else
            printf(
"No\n");
    }
    
return 0;
}
posted @ 2009-05-08 11:28  往往  阅读(179)  评论(0编辑  收藏  举报