Time Limit: 6000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 437    Accepted Submission(s): 157
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define MM(a,b) memset(a,b,sizeof(a));
const double eps = 1e-10;
const int  inf =0x7f7f7f7f;
const double pi=acos(-1);
const int maxn=100+100000;
int deg[maxn];
vector<int> G[maxn];
struct node{
    int id;
    bool operator<(const node &b) const
    {
        return this->id<b.id;
    }
}ne[maxn];
int main()
{
    int cas,n,m;
    scanf("%d",&cas);
    while(cas--)
    {
         scanf("%d %d",&n,&m);
         ll ans=0;
         for(int i=1;i<=n;i++)
         {
             G[i].clear();
             ne[i].id=i;
         }
         memset(deg,0,sizeof(deg));
         priority_queue<node> q;
         for(int i=1;i<=m;i++)
         {
             int u,v;
             scanf("%d %d",&u,&v);
             G[u].push_back(v);
             deg[v]++;
         }
         for(int i=1;i<=n;i++) if(!deg[i]) q.push(ne[i]);
         int minn=n+1;
         while(q.size())
         {
             node k=q.top();q.pop();
             int u=k.id;
             minn=min(minn,u);
             ans+=minn;
             for(int j=0;j<G[u].size();j++)
             {
                 int v=G[u][j];
                 deg[v]--;
                 if(!deg[v]) q.push(ne[v]);
             }
         }
         printf("%lld\n",ans);
    }
    return 0;
}
分析:很好的一道题,,题目中A B意味着要想选B得必须先选A,那么我们可以由A向B连接一条边
,最后构建了图后,考虑选第一个点时,入度为不为0的点是肯定不能选的,那么我们只能在入度入度为0
的点中选,根据贪心的原则,我们选编号最大的作为第一个点(优先队列维护),然后再删除与之相连结的边,再将新出现的入度位0的点压入优先队列。
原文:http://www.cnblogs.com/smilesundream/p/5515658.html