Luogu1613 跑路-倍增+Floyd

Solution

挺有趣的一道题， 仔细想想才想出来

先用$mp[i][j][dis]$ 是否存在一条 $i$ 到 $j$ 的长度为 $2^{dis}$ 的路径。

转移 ：

for (int dis = 1; dis < base; ++dis)
        for (int k = 1; k <= n; ++k)
            for (int i = 1; i <= n; ++i) if (mp[i][k][dis - 1])
                for (int j = 1; j <= n; ++j) if (mp[k][j][dis - 1])
                    mp[i][j][dis] = 1;

 

若$mp[i][j][dis] = 1$, 则把 $f[i][j]$ 记为$1$

然后再用$f[i][j]$ 去跑$Floyd$。 这样找出的路径 一定是最短的（因为能合成 $2^dis$ 的路径都已经被记录了

 

Code

#include<cstdio>
#include<cstring>
#include<algorithm>
#define rd read()
using namespace std;

const int N = 55;
const int base = 32;

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

int read() {
    int X = 0, p = 1; char c = getchar();
    for (; c > '9' || c < '0'; c = getchar())
        if (c == '-') p = -1;
    for (; c >= '0' && c <= '9'; c = getchar())
        X = X * 10 + c - '0';
    return X * p;
}

void cmin(int &A, int B) {
    if (A > B) A = B;
}

int main()
{
    n = rd; m = rd;
    memset(f, 63, sizeof(f));
    for (int i = 1; i <= m; ++i) {
        int u = rd, v = rd;
        mp[u][v][0] = 1;
    }
    for (int dis = 1; dis < base; ++dis)
        for (int k = 1; k <= n; ++k)
            for (int i = 1; i <= n; ++i) if (mp[i][k][dis - 1])
                for (int j = 1; j <= n; ++j) if (mp[k][j][dis - 1])
                    mp[i][j][dis] = 1;
    for (int dis = 0; dis < base; ++dis)
        for (int i = 1; i <= n; ++i)
            for (int j = 1; j <= n; ++j) if (mp[i][j][dis])
                f[i][j] = 1;
    for (int k = 1; k <= n; ++k)
        for (int i = 1; i <= n; ++i)
            for (int j = 1; j <= n; ++j)
                cmin(f[i][j], f[i][k] + f[k][j]);
    printf("%d\n", f[1][n]);
}