Reward

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1072????Accepted Submission(s): 318

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.

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Sample Input

2 1
1 2
2 2
1 2
2 1

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Sample Output

1777
-1

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?????? 最经典的拓扑排序，题目大意：给你a和b，则a的功劳大于b的功劳，即a所得的报酬比要b的多，先输入n和m，表示有n个人，m种关系。报酬至少是888元，问你要分配这些报酬，至少要给多少钱？如果这些关系出现问题（出现环），输出-1。

??????? 就是一个拓扑排序，不解释了，直接代码吧。

链接：http://acm.hdu.edu.cn/showproblem.php?pid=2647

代码:

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

const int N = 10005;
const int M = 20005;

struct edge
{
    int w;
    edge *next;
}*e[N];

edge memory[M];
int cnt;
int deg[N], money[N];
int n, m;

void init()     //初始化
{
    cnt = 0;
    for(int i = 0; i <= n; i++)
    {
        e[i] = NULL;
        money[i] = 888;
        deg[i] = 0;
    }
}

void add(int x, int y)  //建边
{
    edge *p = &memory[cnt++];
    p->w = y;
    p->next = e[x];
    e[x] = p;
}

int main()
{
    int i, x, y, k, u, num, sum;
    while(scanf("%d %d", &n, &m) != EOF)
    {
        init();
        sum = 0;
        num = n;
        for(i = 1; i <= m; i++)
        {
            scanf("%d %d", &x, &y);
            add(y, x);  //建边
            deg[x]++;   //度数++
        }
        /// 拓扑排序
        queue<int> Q;
        for(i = 1; i <= n; i++)
        {
            if(deg[i] == 0) Q.push(i);
        }
        while(!Q.empty())
        {
            k = Q.front();
            Q.pop();
            num--;
            for(edge *p = e[k]; p; p = p->next) //链表代替矩阵
            {
                u = p->w;
                if(--deg[u] == 0)   //和k连接的点度数--，若为0，入队列
                {
                    money[u] = money[k] + 1;    //钱比连接点k的钱多1
                    Q.push(u);
                }
            }
        }
        if(num > 1) printf("-1\n");     //入队列次数少于n，证明有环
        else
        {
            for(i = 1; i <= n; i++)
            {
                sum += money[i];
            }
            printf("%d\n", sum);
        }
    }

    return 0;
}

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