Reward HDU

Reward

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2647 Accepted Submission(s): 768

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

-1

-1777

题意：

让人恶心的不能在恶心的题。。只想说出题作者你太很了。

该题有几个陷阱需要注意：

一：不能在用简单的拓扑了，因为此时显然超内存。只能用到邻接表。

二：注意多劳多得，可能前面的一个人后面跟着好几个，即，后面那几个人的工资是一样的（本人表示本坑了一个上午）

最后题意简单，就是叫你判读是否有环。

详细过程就自己看代码吧，不懂得欢迎提问。共同进步。

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

const int N = 10005;
struct edge
{
    int w;
    edge *next;
}*e[N];

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 = (edge *)malloc(sizeof(edge));
    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++
        }
        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
");     //入队列次数少于n，证明有环
        else
        {
            for(i = 1; i <= n; i++)
            {
                sum += money[i];
            }
            printf("%d
", sum);
        }
    }

    return 0;
}