HDU 2647（拓扑+邻接表）

最新推荐文章于 2022-05-16 16:31:42 发布

真的好难

最新推荐文章于 2022-05-16 16:31:42 发布

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本文介绍了一个奖励分配问题，工厂老板希望在满足员工需求的同时使用最少的资金分配奖励，每个员工的奖励至少为888。通过使用拓扑排序算法，确定是否可以满足所有需求，并计算最小总成本。

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Reward

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5383 Accepted Submission(s): 1632

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

//a 大于 b 所以b->a 记录a的入度从入度为0的点开始向上money递增

#include <stdio.h>
#include <queue>
#include <string.h>
using namespace std;

struct Node
{
    int v;
    int next;
}Edge[20020];
int pre[10010];
int money[10010];
int in[10010];

int ans,sum;
int n,m;
void tops()
{
    queue <int> q;
    sum=ans=0;
    for(int i=1;i<=n;i++)
        if(in[i]==0)
            q.push(i);
    while(!q.empty())
    {
        int b=q.front();
        q.pop();
        sum+=money[b];
        ans++;
        for(int i=pre[b];i!=-1;i=Edge[i].next)
        {
            in[Edge[i].v]--;
            if(in[Edge[i].v]==0)
            {
                q.push(Edge[i].v);
                money[Edge[i].v]=money[b]+1;
            }
        }
    }

}

int main()
{
    while(~scanf("%d%d",&n,&m))
    {

        for(int i=1;i<=n;i++)
            money[i]=888;
        int a,b,index=1;
        memset(pre,-1,sizeof(pre));
        memset(in,0,sizeof(in));
        for(int i=0;i<m;i++)
        {
            scanf("%d%d",&a,&b);
            Edge[index].v=a;
            Edge[index].next=pre[b];
            pre[b]=index++;
            in[a]++;
        }
        tops();
        if(ans!=n)
            printf("-1\n");
        else
            printf("%d\n",sum);
    }
    return 0;
}