【拓扑排序】HDU 2647 Reward【2012/3/25更新】

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

题意：先给出人数n和关系数m，再给出工人之间的工资大小关系，例如给出a b表示a>b，求老板至少要给多少工资【注：工人至少有888元的工资】

Sample Input
2 1
1 2
2 2
1 2
2 1
6 5
2 4
3 6
3 5
1 2
1 3
4 3
1 2
3 4
2 3

Sample Output
1777
-1
5532
3558

#include <iostream>
#include <queue>
using namespace std;
#define M 10005

struct edge{
    int v, w, nxt;
}e[20005];

int ind[M], p[M], cnt, money[M], n;

void init ()
{
    cnt = 0;
    memset (p, -1, sizeof(p));
    memset (ind, 0, sizeof(ind));
    for (int i = 1; i <= n; i++) money[i] = 888;
}

void addedge (int u, int v)
{
    e[cnt].v = v, e[cnt].nxt = p[u], p[u] = cnt++;
}

int main()
{
    int m, u, v, i, num, ans = 0;
    while (~scanf ("%d%d", &n, &m))
    {
        init ();
        while (m--)
        {
            scanf ("%d%d", &u, &v);
            addedge (v, u);		//逆向思维
            ind[u]++;
        }
        queue<int> q;
        for (i = 1; i <= n; i++)
            if (ind[i] == 0)
                q.push (i);
        num = ans = 0;
        while (!q.empty())
        {
            u = q.front();
            q.pop();
            ans += money[u];	//累加确定工资
            num++;
            for (i = p[u]; i != -1; i = e[i].nxt)
            {
                if (--ind[e[i].v] == 0)
                {
                    money[e[i].v] = money[u] + 1;
                    q.push (e[i].v);
                }
            }
        }
        if (num < n) ans = -1;
        printf ("%d\n", ans);
    }
    return 0;
}