BZOJ 1491 社交网络 Floyd 最短路的数目

题目链接：

https://www.lydsy.com/JudgeOnline/problem.php?id=1491

题目大意：

见链接

思路：

直接用floyd算法求最短路，同时更新最短路的数目即可。

#include<bits/stdc++.h>
#define IOS ios::sync_with_stdio(false);//不可再使用scanf printf
#define Max(a, b) ((a) > (b) ? (a) : (b))//禁用于函数，会超时
#define Min(a, b) ((a) < (b) ? (a) : (b))
#define Mem(a) memset(a, 0, sizeof(a))
#define Dis(x, y, x1, y1) ((x - x1) * (x - x1) + (y - y1) * (y - y1))
#define MID(l, r) ((l) + ((r) - (l)) / 2)
#define lson ((o)<<1)
#define rson ((o)<<1|1)
#define Accepted 0
#pragma comment(linker, "/STACK:102400000,102400000")//栈外挂
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
    while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

typedef long long ll;
const int maxn = 100 + 10;
const int MOD = 1000000007;//const引用更快，宏定义也更快
const int INF = 1e9 + 7;
const double eps = 1e-6;

int Map[maxn][maxn];
ll p[maxn][maxn];
double ans[maxn];
int main()
{
    int n, m;
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++)for(int j = 1; j <= n; j++)Map[i][j] = INF;
    while(m--)
    {
        int u, v, w;
        scanf("%d%d%d", &u, &v, &w);
        Map[u][v] = Map[v][u] = w;
        p[u][v] = p[v][u] = 1;//路径条数
    }
    for(int k = 1; k <= n; k++)
    {
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= n; j++)
            {
                if(Map[i][k] + Map[k][j] < Map[i][j])
                {
                    Map[i][j] = Map[i][k] + Map[k][j];
                    p[i][j] = p[i][k] * p[k][j];
                }
                else if(Map[i][k] + Map[k][j] == Map[i][j])
                    p[i][j] += p[i][k] * p[k][j];
            }
        }
    }
    for(int k = 1; k <= n; k++)//k为s-t的必经点
        for(int s = 1; s <= n; s++)
            for(int t = 1; t <= n; t++)
                if(s != t && s != k && k != t)
    {
        if(Map[s][k] + Map[k][t] == Map[s][t])//最短路径可以经过k点
        {
            ans[k] += 1.0 * p[s][k] * p[k][t] / p[s][t];
        }
    }
    for(int i = 1; i <= n; i++)printf("%.3f
", ans[i]);
    return Accepted;
}