bzoj 1491

思路：先求出每两点之间的最短路，建出n个最短路径图，然后枚举起点终点和中间点，计算条数用到拓扑图dp。。。

看别人的方法很巧妙地用floyd在计算最短路的时候就可以直接计算条数啦。。。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int, int>

using namespace std;

const int N = 100 + 7;
const int M = 1e4 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 +7;

int n, m, tot, d[N][N], head[N];
LL dp[N][N];
double ans[N];
bool in[N];
vector<int> Edge[N][N];

struct EDGE {
    int from, to, w, nx;
} edge[M];

void add(int u, int v, int w) {
    edge[tot].from = u;
    edge[tot].to = v;
    edge[tot].w = w;
    edge[tot].nx = head[u];
    head[u] = tot++;
}

void spfa(int s) {
    memset(in, false, sizeof(in));

    queue<int> que; que.push(s);
    d[s][s] = 0, in[s] = true;


    while(!que.empty()) {
        int u = que.front(); que.pop();
        in[u] = false;
        for(int i = head[u]; ~i; i = edge[i].nx) {
            int v = edge[i].to, w = edge[i].w;
            if(d[s][u] + w < d[s][v]) {
                d[s][v] = d[s][u] + w;
                if(!in[v]) in[v] = true, que.push(v);
            }
        }

    }

    for(int i = 0; i < tot; i++) {
        int u = edge[i].from, v = edge[i].to, w = edge[i].w;
        if(d[s][u] + w == d[s][v]) {
            Edge[s][v].push_back(u);
        }
    }
}


LL dfs(int s, int u) {
    if(s == u) return 1;
    if(dp[s][u] != -1) return dp[s][u];
    dp[s][u] = 0;
    for(int k = 0; k < Edge[s][u].size(); k++) {
        int v = Edge[s][u][k];
        dp[s][u] += dfs(s, v);
    }
    return dp[s][u];
}

int main() {
    memset(head, -1, sizeof(head));
    memset(d, inf, sizeof(d));
    memset(dp, -1, sizeof(dp));

    scanf("%d%d", &n, &m);

    for(int i = 1; i <= m; i++) {
        int u, v, w;
        scanf("%d%d%d", &u, &v, &w);
        add(u, v, w); add(v, u, w);
    }

    for(int i = 1; i <= n; i++) {
        spfa(i);
    }

    for(int i = 1; i <= n; i++) {
        for(int j = i + 1; j <= n; j++) {
            LL cnt = dfs(i, j);
            for(int k = 1; k <= n; k++) {
                if(k == i || k == j) continue;
                if(d[i][k] + d[j][k] != d[i][j]) continue;
                LL ret = dfs(i, k) * dfs(j, k);
                ans[k] += 1.0 * ret / cnt;
            }
        }
    }

    for(int i = 1; i <= n; i++) {
        printf("%.3f\n", ans[i] * 2);
    }
    return 0;
}


/*
3
3 2
*/