hdu 2647 Reward

拓扑排序。每次找到入度为0的点存下来。一次找完后，对这些点计算费用。

#include<cstdio>

#include<cstring>

#include<cmath>

#include<vector>

#include<algorithm>

using namespace std;

vector<int>abc[10005];

int rudu[10005];

int linshi[10005];



int main()

{

    int n, m, i, x, y, k;

    while (~scanf("%d%d", &n, &m))

    {

        int sum = 0;

        for (i = 0; i < 10003; i++) abc[i].clear();

        memset(rudu, 0, sizeof(rudu));

        for (i = 0; i < m; i++)

        {

            scanf("%d%d", &x, &y);

            rudu[x]++;

            abc[y].push_back(x);

        }

        int fei = 0;

        int ans = 0;

        while (1)

        {

            if (sum == n) break;

            int uuu = 0;//标记有没有找到

            int yy = 0;

            for (i = 1; i <= n; i++)

            {

                if (rudu[i] == 0)

                {

                    rudu[i]--;

                    uuu = 1;//找到了

                    sum++;

                    linshi[yy] = i; yy++;

                }

            }

            if (uuu == 0) break;

            for (i = 0; i < yy; i++)

            {

                ans = ans + fei;

                for (k = 0; k < abc[linshi[i]].size(); k++) rudu[abc[linshi[i]][k]]--;

            }

            fei++;

        }

        if (sum == n)

        {

            ans = ans + n * 888;

            printf("%d\n", ans);

        }

        else if (sum != n) printf("-1\n");

    }

    return 0;

}