Description
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
有向图 叫你判断给定点有没有参与构成环
因为是有向图
所以并查集不能解决
直接dfs搜一波
#include<stdio.h> #include<iostream> #include<vector> #include <cstring> #include <stack> #include <cstdio> #include <cmath> #include <queue> #include <algorithm> #include <vector> #include <set> #include <map> #include<string> #include<string.h> #include<math.h> typedef long long LL; using namespace std; #define max_v 105 int vis[max_v]; int G[max_v][max_v]; int flag; void dfs(int k) { for(int i=1;i<max_v;i++) { if(vis[i]==1&&G[k][i]==1) { flag=0; return ; } if(G[k][i]==1) { vis[i]=1; dfs(i); vis[i]=0; } } } int main() { int n; int x,y; while(~scanf("%d",&n)) { if(n<0) break; memset(vis,0,sizeof(vis)); memset(G,0,sizeof(G)); for(int i=0;i<n;i++) { scanf("%d %d",&x,&y); if(x!=y) G[x][y]=1; } int k; scanf("%d",&k); vis[k]=1; flag=1; dfs(k); if(flag) printf("Yes "); else printf("No "); } } /* 题目意思: 有向图 叫你判断给定点有没有参与构成环 分析: 因为是有向图 所以并查集不能解决 直接dfs搜一波 */