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
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.
TO MAKE IT SIMPLE, we give every one a number (0, 1, 2,..., N-1). We use their numbers instead of their names.
If it is legal, output "YES", otherwise "NO".
0 1
1 2
2 2
0 1
1 0
0 0
NO
#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 ;
}