HDU4328 Cut the cake（动规：最大子矩形问题/悬线法）

题目链接：传送门

题目大意：

　　给出N*M的字符矩阵（由字符B/R组成），求符合下图条件的子矩阵的最大周长。

　　1 ≤ N，M ≤ 1000。

思路：

　　悬线法。

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;
const int MAX_N = 1e3 + ;

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 = ; i <= N; i++) {
        for (int j = ; j <= M; j++)
            if (j >  && mat[i][j] == mat[i][j-])
                lef[i][j] = lef[i][j-] + ;
            else
                lef[i][j] = ;
        for (int j = M; j >= ; j--)
            if (j < M && mat[i][j] == mat[i][j+])
                rig[i][j] = rig[i][j+] + ;
            else
                rig[i][j] = ;
    }
}

void init2()
{
    for (int i = ; i <= N; i++) {
        for (int j = ; j <= M; j++)
            if (j >  && mat[i][j] != mat[i][j-])
                lef[i][j] = lef[i][j-] + ;
            else
                lef[i][j] = ;
        for (int j = M; j >= ; j--)
            if (j < M && mat[i][j] != mat[i][j+])
                rig[i][j] = rig[i][j+] + ;
            else
                rig[i][j] = ;
    }
}

int dp1()
{
    int ans = ;
    for (int i = ; i <= N; i++) {
        for (int j = ; j <= M; j++) {
            if (i >  && mat[i][j] == mat[i-][j]) {
                up[i][j] = up[i-][j] + ;
                lef[i][j] = min(lef[i][j], lef[i-][j]);
                rig[i][j] = min(rig[i][j], rig[i-][j]);
            }
            else
                up[i][j] = ;
            int len = lef[i][j] + rig[i][j] - ;
            int high = up[i][j];
            ans = max(ans, *len+*high);
        }
    }
    return ans;
}

int dp2()
{
    int ans = ;
    for (int i = ; i <= N; i++) {
        for (int j = ; j <= M; j++) {
            if (i >  && mat[i][j] != mat[i-][j]) {
                up[i][j] = up[i-][j] + ;
                lef[i][j] = min(lef[i][j], lef[i-][j]);
                rig[i][j] = min(rig[i][j], rig[i-][j]);
            }
            else
                up[i][j] = ;
            int len = lef[i][j] + rig[i][j] - ;
            int high = up[i][j];
            ans = max(ans, *len + *high);
        }
    }
    return ans;
}

int main()
{
    int T;
    int kase = ;
    cin >> T;
    while (T--) {
        cin >> N >> M;
        for (int i = ; i <= N; i++)
            for (int j = ; j <= M; j++)
                cin >> mat[i][j];
        int ans = ;
        init1();
        ans = max(ans, dp1());
        init2();
        ans = max(ans, dp2());
        printf("Case #%d: %d\n", kase++, ans);
    }
    return ;
}
/*
2
3 3
BBR
RBB
BBB
1 1
B
*/