Solution

挺有趣的一道題， 仔細想想才想出來

先用$mp[i][j][dis]$ 是否存在一條 $i$ 到 $j$ 的長度為 $2^{dis}$ 的路徑。

轉移 ：

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

若$mp[i][j][dis] = 1$, 則把 $f[i][j]$ 記為$1$

然後再用$f[i][j]$ 去跑$Floyd$。 這樣找出的路徑 一定是最短的（因為能合成 $2^dis$ 的路徑都已經被記錄了

Code

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #define rd read()
 5 using
 
 namespace std;
 6 
 7 const int N = 55;
 8 const int base = 32;
 9 
10 int mp[N][N][N], f[N][N];
11 int n, m;
12 
13 int read() {
14     int X = 0, p = 1; char c = getchar();
15     for (; c > ‘9‘ || c < ‘0‘; c = getchar())
16         if (c == ‘-‘) p = -1;
17     for (; c >= ‘0‘ && c <= ‘
 
9‘; c = getchar())
18         X = X * 10 + c - ‘0‘;
19     return X * p;
20 }
21 
22 void cmin(int &A, int B) {
23     if (A > B) A = B;
24 }
25 
26 int main()
27 {
28     n = rd; m = rd;
29     memset(f, 63, sizeof(f));
30     for (int i = 1; i <= m; ++i) {
31         int u = rd, v = rd;
32         mp[u][v][0] = 1;
33     }
34     for (int dis = 1; dis < base; ++dis)
35         for (int k = 1; k <= n; ++k)
36             for (int i = 1; i <= n; ++i) if (mp[i][k][dis - 1])
37                 for (int j = 1; j <= n; ++j) if (mp[k][j][dis - 1])
38                     mp[i][j][dis] = 1;
39     for (int dis = 0; dis < base; ++dis)
40         for (int i = 1; i <= n; ++i)
41             for (int j = 1; j <= n; ++j) if (mp[i][j][dis])
42                 f[i][j] = 1;
43     for (int k = 1; k <= n; ++k)
44         for (int i = 1; i <= n; ++i)
45             for (int j = 1; j <= n; ++j)
46                 cmin(f[i][j], f[i][k] + f[k][j]);
47     printf("%d\n", f[1][n]);
48 }

Luogu1613 跑路-倍增+Floyd