Hdu4328 cut the cake (motion gauge: Maximum subrectangle problem/suspension method)

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Question:

The N * m character matrix (consisting of character B/R) is given, and the maximum perimeter of the Child matrix that meets the conditions is obtained.

1 ≤ n, m ≤ 1000.

Ideas:

Suspension method.

#include <iostream>#include <cstring>#include <cstdio>using namespace std;const int MAX_N = 1e3 + 5;int N, M;char mat[MAX_N][MAX_N];int lef[MAX_N][MAX_N], rig[MAX_N][MAX_N], up[MAX_N][MAX_N];void init1(){    for (int i = 1; i <= N; i++) {        for (int j = 1; j <= M; j++)            if (j > 1 && mat[i][j] == mat[i][j-1])                lef[i][j] = lef[i][j-1] + 1;            else                lef[i][j] = 1;        for (int j = M; j >= 1; j--)            if (j < M && mat[i][j] == mat[i][j+1])                rig[i][j] = rig[i][j+1] + 1;            else                rig[i][j] = 1;    }}void init2(){    for (int i = 1; i <= N; i++) {        for (int j = 1; j <= M; j++)            if (j > 1 && mat[i][j] != mat[i][j-1])                lef[i][j] = lef[i][j-1] + 1;            else                lef[i][j] = 1;        for (int j = M; j >= 1; j--)            if (j < M && mat[i][j] != mat[i][j+1])                rig[i][j] = rig[i][j+1] + 1;            else                rig[i][j] = 1;    }}int dp1(){    int ans = 0;    for (int i = 1; i <= N; i++) {        for (int j = 1; j <= M; j++) {            if (i > 1 && mat[i][j] == mat[i-1][j]) {                up[i][j] = up[i-1][j] + 1;                lef[i][j] = min(lef[i][j], lef[i-1][j]);                rig[i][j] = min(rig[i][j], rig[i-1][j]);            }            else                up[i][j] = 1;            int len = lef[i][j] + rig[i][j] - 1;            int high = up[i][j];            ans = max(ans, 2*len+2*high);        }    }    return ans;}int dp2(){    int ans = 0;    for (int i = 1; i <= N; i++) {        for (int j = 1; j <= M; j++) {            if (i > 1 && mat[i][j] != mat[i-1][j]) {                up[i][j] = up[i-1][j] + 1;                lef[i][j] = min(lef[i][j], lef[i-1][j]);                rig[i][j] = min(rig[i][j], rig[i-1][j]);            }            else                up[i][j] = 1;            int len = lef[i][j] + rig[i][j] - 1;            int high = up[i][j];            ans = max(ans, 2*len + 2*high);        }    }    return ans;}int main(){    int T;    int kase = 1;    cin >> T;    while (T--) {        cin >> N >> M;        for (int i = 1; i <= N; i++)            for (int j = 1; j <= M; j++)                cin >> mat[i][j];        int ans = 1;        init1();        ans = max(ans, dp1());        init2();        ans = max(ans, dp2());        printf("Case #%d: %d\n", kase++, ans);    }    return 0;}/*23 3BBRRBBBBB1 1B*/

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Hdu4328 cut the cake (motion gauge: Maximum subrectangle problem/suspension method)