HDU2647Reward （拓扑排序）

Reward

Description

Dandelion's uncle is a boss of a factory. As the spring festival is coming , he wants to distribute rewards to his workers. Now he has a trouble about how to distribute the rewards. 
The workers will compare their rewards ,and some one may have demands of the distributing of rewards ,just like a's reward should more than b's.Dandelion's unclue wants to fulfill all the demands, of course ,he wants to use the least money.Every work's reward will be at least 888 , because it's a lucky number.

Input

One line with two integers n and m ,stands for the number of works and the number of demands .(n<=10000,m<=20000) 
then m lines ,each line contains two integers a and b ,stands for a's reward should be more than b's.

Output

For every case ,print the least money dandelion 's uncle needs to distribute .If it's impossible to fulfill all the works' demands ,print -1.

Sample Input

2 1
1 2
2 2
1 2
2 1

Sample Output

1777
-1

思路，构建反图，将工资高的放在后面，好比树，越深的工资越高，这里不算u->v,而是v->u;

无解当前仅当形成环，n个节点没有走完一遍。

代码：

#include <iostream>
#include<vector>
#include<queue>
using namespace std;
const int M=100005;
vector<int>G[M];
int in[M],num[M];
void topsort(int n){
    int ans=0;
    int cnt=0;//统计是否形成环，n个点全部输出则不可能形成环
    queue<int>que;
    //将入度为0 的压入队列
    for (int i = 1; i <= n; i++){
        if(in[i]==0)que.push(i);
    }
    while (!que.empty()){
        cnt++;
        int u=que.front();
        que.pop();
        ans+=num[u];
        for (int i = 0; i <G[u].size(); i++){
            int v = G[u][i];
            if(--in[v]==0)que.push(v);
            num[v]=num[u]+1;
        }
    }
    if(cnt!=n)
        cout<<-1<<endl;
    else{
        cout<<ans<<endl;
    }
}

int main(){
    int n,m,u,v;
    while (cin>>n>>m){
        //初始化
        for (int i = 1; i <= n; i++){
            G[i].clear();
            num[i]=888;
            in[i]=0;
        }
        while (m--){
            cin>>u>>v;
            G[v].push_back(u);
            in[u]++;
        }
        topsort(n);
    }
}