(简单) HDU 5154 Harry and Magical Computer，图论。

Description

　　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.

 

　　题目大致就是说对一个有向图判断是否有环。BC 25 的A题，这场比赛很幸运的爆零了，我觉得是因为A题不停的写错导致整个人都乱了，看来以后要注意，如果第一个题就乱了的话要注意调整，不能影响之后的发挥。

　　这个题目的题解是用的floyd写的，我使用dfs写的，貌似复杂度低一点。

 

代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

bool vis[];
bool rem[];
int n,m;
int first[];
int next[];
int bs[],be[];

bool dfs(int x)
{
    int t=first[x];

    while(t)
    {
        if(rem[be[t]])                //这里要注意先判断。
            return ;

        if(vis[be[t]]==)
        {
            vis[be[t]]=;
            rem[be[t]]=;

            if(dfs(be[t])==)
                return ;
            rem[be[t]]=;
        }

        t=next[t];                //不然会无限循环。
    }    

    return ;
}

bool getans()
{
    for(int i=;i<=n;++i)
        if(vis[i]==)
        {
            memset(rem,,sizeof(rem));        //这里要注意清零。
            rem[i]=;
            vis[i]=;
            if(dfs(i)==)
                return ;
        }

    return ;
}

int main()
{
    int a,b;
    int temp;

    while(~scanf("%d %d",&n,&m))
    {
        memset(vis,,sizeof(vis));
        memset(first,,sizeof(first));
        memset(next,,sizeof(next));

        for(int i=;i<=m;++i)
        {
            scanf("%d %d",&a,&b);
            bs[i]=b;
            be[i]=a;
            temp=first[b];
            first[b]=i;
            next[i]=temp;
        }

        if(getans())
            cout<<"YES"<<endl;
        else
            cout<<"NO"<<endl;
    }

    return ;
}