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最新推荐文章于 2021-10-12 15:02:43 发布

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最新推荐文章于 2021-10-12 15:02:43 发布

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原文链接：http://www.cnblogs.com/zufezzt/p/4437521.html

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;
}