Luogu 1169 [ZJOI2007]棋盘制作 - 动态规划+单调栈
Description
给一个01矩阵, 求出最大的01交错的正方形和最大的01交错的矩阵
Solution
用动态规划求出最大的正方形, 用单调栈求出最大的矩阵。
在这里仅介绍求出最大正方形(求最大矩阵 = 单调栈裸题 传送门 : 不会单调栈的同学可以去学
定义数组$f[ i ][ j ]$ 为以$(i, j) $为右下角的正方形的边长
$up[ i ][ j ]$ 表示从点$(i, j)$往上 $01$交错的长度
$lef[ i ][ j ]$ 表示从点$(i, j)$往右$01$交错的长度
当 $a[ i ][ j ] != a[i - 1][j - 1]$时才可由上一个正方形继续拓展, 否则长度 $f[ i ][ j ] = 1$
于是有转移方程:
$f[ i ][ j ] = 1$ $a[ i ][ j ] == a[i - 1][j - 1]$
$f[ i ][ j ] = min(f[ i - 1][ j - 1] + 1, lef[ i ][ j ], up[ i ][ j ])$ $ a[ i ][ j ] != a[i - 1][ j ]$
Code
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #define rd read() 5 #define rep(i,a,b) for(register int i = (a); i <= (b); i++) 6 #define per(i,a,b) for(register int i = (a); i >= (b); --i) 7 #define R register 8 using namespace std; 9 10 const int N = 3e3; 11 12 int n, m, a[N][N], pre[N], nxt[N], h[N], ans2; 13 int lf[N][N], u[N][N], f[N][N], ans1; 14 int st[N], tp; 15 16 int read() { 17 int X = 0, p = 1; char c = getchar(); 18 for(; c > '9' || c < '0'; c = getchar()) if(c == '-') p = -1; 19 for(; c >= '0' && c <= '9'; c = getchar()) X = X * 10 + c - '0'; 20 return X * p; 21 } 22 23 void work(int x) { 24 rep(i, 1, m) pre[i] = 0, nxt[i] = m + 1; 25 rep(i, 1, m) 26 if(a[x][i] != a[x - 1][i]) h[i]++; 27 else h[i] = 1; 28 tp = 0; 29 rep(i, 1, m) { 30 int pr = 0; 31 while(tp) { 32 if(h[st[tp]] >= h[i]) tp--; 33 else { 34 pr = st[tp]; break; 35 } 36 } 37 pre[i] = pr; 38 st[++tp] = i; 39 } 40 tp = 0; 41 per(i, m, 1) { 42 int nt = m + 1; 43 while(tp) { 44 if(h[st[tp]] >= h[i]) tp--; 45 else { 46 nt = st[tp]; break; 47 } 48 } 49 nxt[i] = nt; 50 st[++tp] = i; 51 } 52 rep(i, 1, m) { 53 int nt = nxt[i], pr = pre[i]; 54 rep(j, i + 1, nt) 55 if(a[x][j] == a[x][j - 1]) {nt = j; break;} 56 per(j, i - 1, pr) 57 if(a[x][j] == a[x][j + 1]) {pr = j; break;} 58 ans2 = max(ans2, (nt - pr - 1) * h[i]); 59 } 60 } 61 62 int main() 63 { 64 n = rd; m = rd; 65 memset(a, -1, sizeof(a)); 66 rep(i, 1, n) rep(j, 1, m) a[i][j] = rd; 67 rep(i, 1, n) rep(j, 1, m) { 68 f[i][j] = 1; 69 if(a[i][j] != a[i][j - 1]) 70 lf[i][j] = lf[i][j - 1] + 1; 71 else lf[i][j] = 1; 72 if(a[i][j] != a[i - 1][j]) 73 u[i][j] = u[i - 1][j] + 1; 74 else u[i][j] = 1; 75 if(a[i][j] != a[i - 1][j - 1]) continue; 76 f[i][j] = min(u[i][j], lf[i][j]); 77 f[i][j] = min(f[i - 1][j - 1] + 1, f[i][j]); 78 ans1 = max(ans1, f[i][j]); 79 } 80 printf("%d\n", ans1 * ans1); 81 rep(i, 1, n) work(i); 82 printf("%d\n", ans2); 83 }