社交网络

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题目大意: 社交网络

看到数据范围果断想到弗洛伊德

但是有如下坑点:

  • 统计答案的时候一定要判断当前点是不是两断点对最短路...

  • 一个数组记录最短路,一个数组记录方案数目.

其中\(f_{i,j}\)表示从i到j的最短路对方案数目,\(g_{i,j}\)表示i到j对最短距离.

然后依题跑一趟就可以了.

代码如下: 

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#include <bits/stdc++.h>
#define int long long
#define ll long long
using namespace std;

const int N = 1e2 + 66;

int n, m;
int g[N][N], f[N][N];

inline void pres_dou()
{
    int i, j; memset(g, 0x3f, sizeof g);
    for (i = 1; i <= n; ++ i) g[i][i] = 0;
    for (i = 1; i <= n; ++ i) for (j = 1; j <= n; ++ j) f[i][j] = 1;
    return;
}

signed main()
{
    int i, j, k, x, y, z;
    n = read(), m = read(), pres_dou();
    for (i = 1; i <= m; ++ i)
    {
        x = read(), y = read(), z = read();
        g[x][y] = g[y][x] = z;
    }

    for (k = 1; k <= n; ++ k) for (i = 1; i <= n; ++ i) for (j = 1; j <= n; ++ j)
    {
        if (i != k && i != j && k != j)
        {
            if (g[i][j] >= g[i][k] + g[k][j])
            {
                if (g[i][j] == g[i][k] + g[k][j]) f[i][j] += f[i][k] * f[k][j];
                else
                {
                    g[i][j] = g[i][k] + g[k][j];
                    f[i][j] = f[i][k] * f[k][j];
                }
            }
        }
    }

    for (k = 1; k <= n; ++ k)
    {
        double res(0);
        for (i = 1; i <= n; ++ i)
        {
            for (j = 1; j <= n; ++ j)
            {
                if (i == k || j == k || ! f[i][j]) continue;
                if (g[i][k] + g[k][j] == g[i][j]) res += 1.0 * f[i][k] * f[k][j] / f[i][j];
            }
        }
        printf ("%.3lf\n", res);
    }

    return 0;
}

后记:

考场代码(30pts)距离AC只差两句话对距离.

弗洛伊德不需要用到inf,少用inf