(簡單) HDU 5154 Harry and Magical Computer，圖論。

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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. 　　題目大致就是說對一個有向圖判斷是否有環。BC 25 的A題，這場比賽很幸運的爆零了，我覺得是因為A題不停的寫錯導致整個人都亂了，看來以後要注意，如果第一個題就亂了的話要注意調整，不能影響之後的發揮。　　這個題目的題解是用的floyd寫的，我使用dfs寫的，貌似複雜度低一點。 代碼如下：

#include<iostream>#include<cstdio>#include<cstring>using namespace std;bool vis[105];bool rem[105];int n,m;int first[105];int next[10005];int bs[10005],be[10005];bool dfs(int x){    int t=first[x];        while(t)    {        if(rem[be[t]])                //這裡要注意先判斷。            return 0;        if(vis[be[t]]==0)        {            vis[be[t]]=1;            rem[be[t]]=1;                        if(dfs(be[t])==0)                return 0;            rem[be[t]]=0;        }        t=next[t];                //不然會無限迴圈。    }        return 1;}bool getans(){    for(int i=1;i<=n;++i)        if(vis[i]==0)        {            memset(rem,0,sizeof(rem));        //這裡要注意清零。            rem[i]=1;            vis[i]=1;            if(dfs(i)==0)                return 0;        }    return 1;}int main(){    int a,b;    int temp;    while(~scanf("%d %d",&n,&m))    {        memset(vis,0,sizeof(vis));        memset(first,0,sizeof(first));        memset(next,0,sizeof(next));                for(int i=1;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 0;}

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(簡單) HDU 5154 Harry and Magical Computer，圖論。