hdu 5154 Harry and Magical Computer 拓扑排序

Harry and Magical Computer

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)


[align=left]Problem Description[/align]
In reward of being yearly outstanding magic student, Harry gets a magical computer. When the computer begins to deal with a process, it will work until the ending of the processes. One day the computer got n processes to deal with. We number the processes from 1 to n. However there are some dependencies between some processes. When there exists a dependencies (a, b), it means process b must be finished before process a. By knowing all the m dependencies, Harry wants to know if the computer can finish all the n processes.

[align=left]Input[/align]
There are several test cases, you should process to the end of file.
For each test case, there are two numbers n m on the first line, indicates the number processes and the number of dependencies. 1≤n≤100,1≤m≤10000
The next following m lines, each line contains two numbers a b, indicates a dependencies (a, b). 1≤a,b≤n

[align=left]Output[/align]
Output one line for each test case.
If the computer can finish all the process print "YES" (Without quotes).
Else print "NO" (Without quotes).

[align=left]Sample Input[/align]

3 2
3 1
2 1
3 3
3 2
2 1
1 3

[align=left]Sample Output[/align]

YES
NO

[align=left]Source[/align]
BestCoder Round #25
题意：有向图判环；
思路：拓扑完没点；

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define mod 1000000007
#define esp 0.00000000001
const int N=2e3+10,M=1e6+10,inf=1e9;
int n,m;
vector<int>edge
;
int du
;
int main()
{
while(~scanf("%d%d",&n,&m))
{
queue<int>q;
memset(du,0,sizeof(du));
for(int i=0;i<=n;i++)
edge[i].clear();
int ans=0;
for(int i=1;i<=m;i++)
{
int u,v;
scanf("%d%d",&u,&v);
edge[u].push_back(v);
du[v]++;
}
for(int i=1;i<=n;i++)
{
if(!du[i])q.push(i);
}
while(!q.empty())
{
int v=q.front();
q.pop();
ans++;
for(int i=0;i<edge[v].size();i++)
{
du[edge[v][i]]--;
if(!du[edge[v][i]])
q.push(edge[v][i]);
}
}
if(ans==n)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}