hdu 2647 Reward（拓扑排序)

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#hdu #拓扑排序

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解决一个工厂老板如何用最少的资金满足员工间奖励金额的相对大小需求的问题，确保每位员工至少获得888元，并通过拓扑排序确定最终的资金分配方案。

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题目：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=23269

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

让最小的人收到888money，比他大的加1，按照拓扑排序增长，最后求出总资金。

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn=2e4+5,N=maxn/2;
int head[N],indeg[N],n,m,ip;
struct node{
    int to,next;
}edge[maxn];
int money[maxn];
void init(){
    ip=0;
    memset(money,0,sizeof(money));
    memset(head,-1,sizeof(head));
    memset(indeg,0,sizeof(indeg));
}
bool topo(){
    int que[maxn],iq=0;
    for(int i=1;i<=n;i++) if(indeg[i]==0){ que[iq++]=i;  money[i]=888; } //cout<<i<<endl; }
    for(int i=0;i<iq;i++){
         for(int k=head[que[i]];k>-1;k=edge[k].next){
             indeg[edge[k].to]--;
             if(indeg[edge[k].to]==0){
                 que[iq++]=edge[k].to; //que里存储的是节点，head存储的是边的编号
                 money[edge[k].to]=money[que[i]]+1;
             }
         }
    }
    if(iq==n) return 1;
    return 0;
}
void addedge(int u,int v){
    edge[ip].to=v;
    edge[ip].next=head[u];
    head[u]=ip++;
    indeg[v]++;
}
int main()
{
    //freopen("cin.txt","r",stdin);
    int a,b;
    while(cin>>n>>m){
        init();
        while(m--){
            scanf("%d %d",&a,&b);
            addedge(b,a);
        }
        int pj=topo();
        if(pj==0){  printf("-1\n");  continue; }
        int ans=0;
        for(int i=1;i<=n;i++){
            ans+=money[i];
        }
        printf("%d\n",ans);
    }
    return 0;
}