http://acm.hdu.edu.cn/showproblem.php?pid=3342
Legal or Not
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5788 Accepted Submission(s): 2678
Problem Description
ACM-DIY is a large QQ group where many excellent acmers get together. It is so harmonious that just like a big family. Every day,many "holy cows" like HH, hh, AC, ZT, lcc, BF, Qinz and so on chat on-line to exchange their ideas. When someone has questions, many warm-hearted cows like Lost will come to help. Then the one being helped will call Lost "master", and Lost will have a nice "prentice". By and by, there are many pairs of "master and prentice". But then problem occurs: there are too many masters and too many prentices, how can we know whether it is legal or not?
We all know a master can have many prentices and a prentice may have a lot of masters too, it's legal. Nevertheless，some cows are not so honest, they hold illegal relationship. Take HH and 3xian for instant, HH is 3xian's master and, at the same time, 3xian is HH's master,which is quite illegal! To avoid this,please help us to judge whether their relationship is legal or not.
Please note that the "master and prentice" relation is transitive. It means that if A is B's master ans B is C's master, then A is C's master.
Input
The input consists of several test cases. For each case, the first line contains two integers, N (members to be tested) and M (relationships to be tested)(2 <= N, M <= 100). Then M lines follow, each contains a pair of (x, y) which means x is y's master and y is x's prentice. The input is terminated by N = 0.
TO MAKE IT SIMPLE, we give every one a number (0, 1, 2,..., N-1). We use their numbers instead of their names.
TO MAKE IT SIMPLE, we give every one a number (0, 1, 2,..., N-1). We use their numbers instead of their names.
Output
For each test case, print in one line the judgement of the messy relationship.
If it is legal, output "YES", otherwise "NO".
If it is legal, output "YES", otherwise "NO".
Sample Input
3 2
0 1
1 2
2 2
0 1
1 0
0 0
0 1
1 2
2 2
0 1
1 0
0 0
Sample Output
YES
NO
NO
题意：给出一个不一定联通的图，判断图中是否有环
题解：典型的拓扑排序判环，我又想到了暴力的dfs但是因为dfs要扫描所有的路径，所以超时了，也可以用强联通分量做
下面是拓扑排序的ac代码
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define N 150
struct Edge{
int to;
int next;
}edge[N];
int head[N];
int Enct;
int in[N];
void init()
{
Enct = ;
memset(head,-,sizeof(head));
memset(in,,sizeof(in));
}
void add(int from , int to )
{
edge[Enct].to = to;
edge[Enct].next = head[from];
head[from]= Enct++;
}
int que[N];
int n;
bool ph()
{
int c = ;
for(int i = ; i < n ;i++)
{
if(in[i]==) que[c++] = i;
}
for(int i = ; i < c; i++)
{
for(int j = head[que[i]] ; j!=-; j= edge[j].next)
{
Edge e = edge[j];
in[e.to]--;
if(in[e.to]==)
que[c++] = e.to;
}
}
//printf("c = %d\n",c);
if(c<n-) return false ;
else return true;
}
int main()
{
int m ;
while(~scanf("%d%d",&n,&m)&&(n!=||m!=))
{
init();
for(int i = ;i < m ;i++)
{
int a , b;
scanf("%d%d",&a,&b);
add(a,b);
in[b]++;
}
if(ph()) printf("YES\n");
else printf("NO\n");
} return ;
}
下面是dfs超时的代码
//这种遍历所有路径的方法一般会超时，真的超时了，嘎嘎
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define N 150
int vis[N];
int n ;
struct Edge{
int to ;
int next;
}edge[N];
int head[N];
int Enct;
void init()
{
Enct = ;
memset(head,-,sizeof(head));
memset(vis,,sizeof(vis));//标记0为未访问
}
void add(int from , int to )
{
edge[Enct].to = to;
edge[Enct].next = head[from];
head[from] = Enct++;
}
/*bool dfs(int i )
{
if(vis[i]) return false;
vis[i] = 1;
printf("vis[%d] = %d\n",i,vis[i]);
for(int j = head[i] ; j!=-1; j = edge[j].next)
{
Edge e = edge[j];
dfs(e.to);
}
return true;
}*/
bool tm = true;
bool dfs(int i )
{
vis[i]=;
for(int j = head[i] ; j!=- ;j = edge[j].next)
{
Edge e = edge[j];
if(vis[e.to]==) tm = false;
else
{
dfs(e.to);
vis[e.to]=;//保证dfs走的是一条链，每次回溯的时候相当于走反向所以标记成未访问
}
}
return tm;
}
int main()
{
int m ;
while(~scanf("%d%d",&n,&m)&&(n!=||m!=))
{
init();
tm = true;
for(int i = ;i < m ;i++)
{
int a ,b;
scanf("%d%d",&a,&b);
add(a,b);
}
bool flag = true;
for(int i= ; i < n ;i++)
{
if(flag == false) break;
if(vis[i]==)
flag = dfs(i);
}
if(flag) printf("YES\n");
else printf("NO\n");
}
return ;
}