HDU-2647 Reward(拓扑排序)

最新推荐文章于 2020-04-30 22:38:18 发布

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原文链接：http://www.cnblogs.com/Pretty9/p/7413231.html

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#java

本文介绍了一个奖励分配问题，工厂老板需要在满足员工需求的同时使用最少的资金为每位员工发放不低于888的奖金。通过构建图模型并运用拓扑排序的方法求解最小花费。

Reward

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9799 Accepted Submission(s): 3131

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

#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>

using namespace std;
const int N = 10000 + 15;
int n, m, in[N], val[N];
vector<int> edge[N];

void Solve_question(){
    int ans = 0, cnt = 0;
    queue <int> Q;

    for(int i = 1; i <= n; i++)
        if(!in[i]) { Q.push(i); val[i] = 888; }
    while(!Q.empty()){
        int u = Q.front(); Q.pop();
        cnt++;
        for(int i = 0; i < (int)edge[u].size(); i++){
            int v = edge[u][i];
            if(--in[v] == 0){
                Q.push(v);
                val[v] = val[u] + 1;
            }
        }
    }
    if(cnt < n) puts("-1");
    else{
        for(int i = 1; i <= n; i++)
            ans += val[i];
        printf("%d\n", ans);
    }
}

void Input_data(){
    for(int i = 1; i <= n; i++) edge[i].clear(), val[i] = in[i] = 0;
    int u, v;
    for(int i = 1; i <= m; i++){
        scanf("%d %d", &u, &v);
        in[u]++;
        edge[v].push_back(u);
    }
}

int main(){
    while(scanf("%d %d", &n, &m) == 2){
        Input_data();
        Solve_question();
    }
}