Reward

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5636 Accepted Submission(s): 1712

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

题意：

老板要给非常多员工发奖金。可是部分员工有个虚伪心态，觉得自己的奖金必须比某些人高才心理平衡。可是老板非常人道。想满足全部人的要求。而且非常吝啬，想画的钱最少，问满足全部人的前提下最少花多少钱。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#define maxn 11000
using namespace std;

char map[maxn];
int indu[maxn];
int head[maxn], cnt;
int n, m;
int a[maxn];

struct node {
    int u, v, next;
};

node edge[110000];

void init(){
    cnt = 0;
    memset(head, -1, sizeof(head));
    memset(indu, 0, sizeof(indu));
    for(int i = 1; i <= n; ++i)//全部人的工资一開始都为888
        a[i] = 888;
}

void add(int u, int v){
    edge[cnt] = {u, v, head[u]};
    head[u] = cnt++;
}

void input(){
    while(m--){
        int a, b;
        scanf("%d%d", &b, &a);
        add(a, b);//注意是反向的, wa了一次
        indu[b]++;
    }
}

void topsort(){
    queue<int >q;
    int sum = 0;
    int ans = 0;
    for(int i = 1; i <= n; ++i){
        if(!indu[i]){
            q.push(i);
            ans++;
        }
    }
    while(!q.empty()){
        int u = q.front();
        sum += a[u];
        q.pop();
        for(int i = head[u]; i != -1; i = edge[i].next){
            int v = edge[i].v;
            indu[v]--;
            if(!indu[v]){
                q.push(v);
                a[v] = a[u] + 1;//保证后一个人的工资比前一个人的工资高
                ans++;
            }
        }
    }
    if(ans == n)
        printf("%d\n", sum);
    else
        printf("-1\n");
}

int main (){
    while(scanf("%d%d", &n, &m) != EOF){
        init();
        input();
        topsort();
    }
    return 0;
}

HDU 2647--Reward【拓扑排序】

Reward

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