Reward(HDOJ-2647)

最新推荐文章于 2022-07-28 10:35:58 发布

原创最新推荐文章于 2022-07-28 10:35:58 发布 · 180 阅读

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本文介绍了一个工厂老板在春节前如何合理分配员工奖励的问题，并提供了一种解决方案，通过拓扑排序来确定最少的资金分配方式，同时确保满足所有员工提出的相对薪酬需求。

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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

题意：

给出n个工人的m条工资对比关系，工人的最低工资是888。

思路：

很明显，只要拓扑排序一遍，得出每个工人的工资高低关系，然后高的人比低的人高1元，用res记录每个人的工资，如果最后统计出的人数小于n说明无法排序，输出-1。

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cstdlib>
#include <map>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <list>
#include <cmath>
#include <algorithm>
#define MST(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int MAXN = 10000 + 10;
int n, m;
vector <int> g[MAXN];
int res[MAXN], dgree[MAXN], ans;
void init() {
    for (int i = 1; i <= n; i++) g[i].clear();
    MST(res, 0);
    MST(dgree, 0);
    ans = 0;
}
bool topo() {
    queue <int> q;
    int pos = 0;
    for (int i = 1; i <= n; i++)
        if (!dgree[i]) {
            q.push(i);
            res[i] = 888;
            pos++;
        }
    while (!q.empty()) {
        int u = q.front();
        q.pop();
        ans += res[u];
        for (int i = 0; i < g[u].size(); i++) {
            if (--dgree[g[u][i]] == 0) {
                res[g[u][i]] = res[u] + 1;
                pos++;
                q.push(g[u][i]);
            }
        }
    }
    if (pos < n) return false;
    return true;
}
int main() {
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    while (cin >> n >> m) {
        init();
        while (m--) {
            int x, y;
            cin >> x >> y;
            g[y].push_back(x);
            dgree[x]++;
        }
        if (!topo()) cout << "-1" << endl;
        else cout << ans << endl;
    }
}