HDU 1505 & amp; POJ 1964 City Game (recursive + scanning)

A matrix. Some grids are F and some are R. You need to find the largest sub-matrix so that all the matrices are F.

Idea: enumerate each sub-Matrix directly, obviously the complexity is O (m ^ 3 n ^ 3), obviously TLE.

We can use scanning method: up (I, j) is used to represent a grid (I, j) Suspension Line (the suspension line refers to the line segment before this grid goes up to the grid of the first R). Left (I, j) indicates the position where the suspension line can be moved to the left by scanning the leftmost position (R will be stopped), right (I, j) indicates the position where the drape line can be moved to the rightmost position (R will stop ). Then, the lattice (I, j) can correspond to a maximum F rectangle, with row I as the bottom edge and row I-up (I, j) + 1 as the top bottom edge, the column where left (I, j) is located is the left column, and the column where right (I, j) is located is the right rectangle.

The final answer is the maximum value in the rectangle corresponding to each grid.

Up (I, j) can be recursive: (1) if map [I] [j] = 'F', up (I, j) = up (I-1, j) + 1 (2) if map [I] [j] = 'R', up (I, j) = 0

Left (I, j) can be recursive: (1) if map [I] [j] = 'F', left (I, j) = max {left (I-1, j), lo + 1} (lo is the column 'R' closest to the left of the grid (I, j) (2) if map [I] [j] = 'R', left (I, j) = 0

Right (I, j) can be recursive: (1) if map [I] [j] = 'F', right (I, j) = max {right (I-1, j), ro-1} (ro is the column of the nearest 'r' on the right of the grid (I, j) (2) if map [I] [j] = 'R ', right (I, j) = n

 

Note: The number of data read from this question is large. scanf will use TLE .... Getchar...

 

# Include <cstdio> # include <iostream> # define MAXN 1005 using namespace std; int T, n, m, righ [MAXN] [MAXN], up [MAXN] [MAXN], lef [MAXN] [MAXN]; char map [MAXN] [MAXN]; int main () {scanf ("% d ", & T); while (T --) {int ans = 0; scanf ("% d", & m, & n); getchar (); for (int I = 0; I <m; ++ I) for (int j = 0; j <n; ++ j) {char ch = getchar (); while (ch! = 'F' & ch! = 'R') ch = getchar (); // process the read. Use while until the read is F or R. map [I] [j] = ch ;} for (int j = 0; j <n; ++ j) for (int I = 0; I <m; ++ I) if (map [I] [j] = 'R') up [I] [j] = 0; else up [I] [j] = I? Up [I-1] [j] +; // recursive up (I, j) for (int I = 0; I <m; ++ I) for (int j = 0, lo =-1; j <n; ++ j) if (map [I] [j] = 'R') lo = j, lef [I] [j] = 0; else lef [I] [j] = I? Max (lo + 1, lef [I-1] [j]): lo + 1; // recursive left (I, j) for (int I = 0; I <m; ++ I) for (int j = n-1, ro = n; j> = 0; -- j) if (map [I] [j] = 'R ') ro = j, righ [I] [j] = n; else righ [I] [j] = I? Min (ro-1, righ [I-1] [j]): ro-1; // recursive right (I, j) for (int I = 0; I <m; ++ I) for (int j = 0; j <n; ++ j) if (map [I] [j] = 'F') ans = max (ans, (righ [I] [j]-lef [I] [j] + 1) * up [I] [j]); // calculate the maximum value of the answer printf ("% d \ n", ans * 3);} return 0 ;}