hdu 2647(拓扑排序)

Reward

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7231 Accepted Submission(s): 2263

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:一个公司有n个人,现在知道了m个关系,输入 a b的话代表 a 的工资要比 b 高,最低工资是 888 元，问老板最后需要发的最少工资?

题解:建立 b - > a的有向图,进行拓扑排序.然后从每一个起始点的工资从888开始，下一个点比上一个点高 1 ，计算即可。

#include <iostream>
#include <cstdio>
#include <string.h>
#include <queue>
#include <algorithm>
#include <math.h>
using namespace std;
const int N = 10005;
const int M = 20005;
struct Edge{
    int v,next;
}edge[M];
int head[N];
int indegree[N];
int w[N];
int tot;
void init(){
    memset(head,-1,sizeof(head));
    memset(indegree,0,sizeof(indegree));
    tot = 0;
}
void addEdge(int u,int v,int& tot){
    edge[tot].v = v,edge[tot].next = head[u],head[u] = tot++;
}
int main(){
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF){
        init();
        for(int i=1;i<=m;i++){
            int u,v;
            scanf("%d%d",&u,&v);
            addEdge(v,u,tot);
            indegree[u]++;
        }
        queue<int> q;
        int tot = 0;
        for(int i=1;i<=n;i++){
            if(!indegree[i]) {
                q.push(i);
                w[i] = 888;
                tot+=w[i];
            }
        }
        int cnt = 0;
        while(!q.empty()){
            int u = q.front();
            q.pop();
            cnt++;
            for(int k = head[u];k!=-1;k=edge[k].next){
                int v = edge[k].v;
                indegree[v]--;
                if(!indegree[v]){
                    w[v] = w[u]+1;
                    tot+=w[v];
                    q.push(v);
                }
            }
        }
        if(cnt!=n) printf("-1
");
        else{
            printf("%d
",tot);
        }
    }
    return 0;
}