HDU 2647 - Reward(拓扑排序)

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Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3001    Accepted Submission(s): 899

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.

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.

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 11 22 21 22 1

Sample Output



#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int MAXN=11111;const int MAXM=21111;const int INF=0x3f3f3f3f;int edge,n,m,a,b;int head[MAXN],ind[MAXN],que[MAXN],cost[MAXN];struct EdgeNode{    int to;    int next;}edges[MAXM];void addedge(int u,int v){    edges[edge].to=v;    edges[edge].next=head[u];    head[u]=edge++;}int main(){    while(~scanf("%d%d",&n,&m))    {        memset(head,-1,sizeof(head));        memset(ind,0,sizeof(ind));        edge=0;        int cnt=0,ans=0;        for(int i=1;i<=n;i++) cost[i]=888;        for(int i=0;i<m;i++)        {            scanf("%d%d",&a,&b);            addedge(b,a);            ind[a]++;        }        for(int i=1;i<=n;i++)        {            if(ind[i]==0)            {                que[cnt++]=i;                ans+=cost[i];            }        }        for(int i=0;i<cnt;i++)        {            for(int j=head[que[i]];j!=-1;j=edges[j].next)            {                ind[edges[j].to]--;                if(ind[edges[j].to]==0)                {                    que[cnt++]=edges[j].to;                    cost[edges[j].to]=cost[que[i]]+1 ;                    ans+=cost[edges[j].to];                }            }        }        if(cnt!=n) printf("-1\n");        else printf("%d\n",ans);    }    return 0;}