HDU2647：Reward（拓扑排序）

最新推荐文章于 2019-07-08 08:47:29 发布

原创最新推荐文章于 2019-07-08 08:47:29 发布 · 249 阅读

0 ·

CC 4.0 BY-SA版权

拓扑排序专栏收录该内容

10 篇文章

订阅专栏

解决一个工厂老板如何用最少的花费满足员工间的相对薪资要求的问题。通过拓扑排序算法，确保每位员工的薪资至少为888，并且符合指定的薪资高低顺序。

Reward

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9472 Accepted Submission(s): 3036

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

Sample Output

  
   1777
-1

Author

dandelion

Source

曾是惊鸿照影来

题意：给N个人发功资，有M个工资关系关系，a比b的工资一定要多，底薪为888，问工资支出总和的最小值，不行就输出-1.

思路：拓扑排序，反向建边，从工资最低的人开始发起，判断下成环的情况即可。

# include <iostream>
# include <cstdio>
# include <cstdlib>
# include <cstring>
# include <vector>
# define pb push_back
using namespace std;
const int maxn = 1e4+30;
int n, a[maxn], in[maxn], q[maxn<<1];
vector<int>v[maxn];
long long ans;
void solve()
{
    int icount = 0;
    ans = 0;
    int l=0, r=0;
    for(int i=1; i<=n; ++i)
        if(in[i] == 0)
            q[r++] = i, a[i] = 888, ans += 888, ++icount;
    if(ans == 0) {ans = -1;return;}
    while(l<r)
    {
        int u = q[l++];
        for(int i=0; i<v[u].size(); ++i)
        {
            if(in[v[u][i]] == 0) if(a[v[u][i]] <= a[u]) {ans = -1;return;}
            if(--in[v[u][i]] == 0)
            {
                a[v[u][i]] = a[u]+1;
                ans += a[v[u][i]];
                ++icount;
                q[r++] = v[u][i];
            }
        }
    }
    if(icount < n) ans = -1;//icount记录已经处理的员工数。
}
int main()
{
    int m;
    while(~scanf("%d%d",&n,&m))
    {
        memset(a, 0, sizeof(a));
        memset(in, 0, sizeof(in));
        for(int i=1; i<=n; ++i) v[i].clear();
        while(m--)
        {
            int a, b;
            scanf("%d%d",&a,&b);
            v[b].pb(a);
            ++in[a];
        }
        solve();
        printf("%lld\n",ans);
    }
    return 0;
}