Reward

Problem 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  

Author dandelion  

Source
曾是惊鸿照影来  

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题目大意：

n个人，m条边，每条边a，b 表示a比b的工资多1，每个人的工资至少888，问你工资和至少多少？如果出现矛盾关系，输出-1

解题思路：

根据人的工资关系建立拓扑图，工资尽量从888开始，然后根据是否能全部排好序判断是出现矛盾关系。

解题思路：

#include <iostream>
#include <cstdio>
#include <queue>
#include <vector>
using namespace std;

const int maxn=11000;
int n,m,ans[maxn],r[maxn];
vector <vector<int> > v;

void input(){
    v.clear();
    v.resize(n+1);
    for(int i=0;i<=n;i++){
        ans[i]=-1;
        r[i]=0;
    }
    int a,b;
    while(m-- >0){
        scanf("%d%d",&a,&b);
        v[b].push_back(a);
        r[a]++;
    }
}

void solve(){
    queue <int> q;
    for(int i=1;i<=n;i++){
        if(r[i]==0) q.push(i);
    }
    int tmp=888;
    while(!q.empty()){
        int qsize=q.size();
        while(qsize-- >0){
            int s=q.front();
            q.pop();
            ans[s]=tmp;
            for(int i=0;i<v[s].size();i++){
                int d=v[s][i];
                r[d]--;
                if(r[d]==0) q.push(d);
            }
        }
        tmp++;
    }
    int sum=0;
    for(int i=1;i<=n;i++){
        if(ans[i]>0) sum+=ans[i];
        else{
            sum=-1;
            break;
        }
    }
    printf("%d\n",sum);
}

int main(){
    while(scanf("%d%d",&n,&m)!=EOF){
        input();
        solve();
    }
    return 0;
}