洛谷P1730 最小密度路径(floyd)

题意

题目链接

sol

zz floyd。

很显然的一个dp方程\(f[i][j][k][l]\)表示从\(i\)到\(j\)经过了\(k\)条边的最小权值

可以证明最优路径的长度一定\(\leqslant n\)

然后一波\(n^4\) dp就完了

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int inf = 1e9 + 10;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int n, m;
int f[51][51][1001];
int main() {
    //memset(f, 0x3f, sizeof(f));
    n = read(); m = read();
    for(int k = 1; k <= n; k++)
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++)
                f[i][j][k] = inf;
    for(int i = 1; i <= m; i++) {
        int x = read(), y = read(), w = read();
        f[x][y][1] = min(f[x][y][1], w);
    }
    for(int l = 2; l <= n; l++)// num of edge 
        for(int k = 1; k <= n; k++) // mid point 
            for(int i= 1; i <= n; i++) // start point 
                for(int j = 1; j <= n; j++) // end point
                    f[i][j][l] = min(f[i][j][l], f[i][k][l - 1] + f[k][j][1]);
    int q = read();
    while(q--) {
        int x = read(), y = read();
        double ans = 1e18;
        for(int i = 1; i <= n; i++) if(f[x][y][i] != inf) ans = min(ans, (double) f[x][y][i] / i);
        if(ans == 1e18) puts("omg!");
        else printf("%.3lf\n", ans);
    }
    return 0;
}
/*
*/