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

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 = ;
 const int base = ;

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

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

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

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