题意：给出一个01矩阵，每次询问一个矩形中的最大全1正方形的边长。
用 dp[i][j] 存储以 (i,j) 为右下角的最大全1正方形，转移方程就是
dp[i][j]={min{dp[i−1][j],dp[i][j−1],dp[i−1][j−1]}+10(val[i][j]=1)(val[i][j]=0) .然后用二维RMQ记录子矩阵的最大正方形的边长。 然后对于询问 (x1,y1),(x2,y2) ，二分边长mid，如果询问的矩形中存边长为mid的全1正方形那么必然是以矩形 (x1+mid−1,y1+mid−1)−(x2,y2) 中的某一个点为右下角，所以直接矩形的最大值check即可。 #include <cstdio> #include <iostream> #include <cstring> #include <queue> #include <cmath> #include <algorithm> #include <stack> #include <map> #pragma comment(linker, "/STACK:102400000,102400000") #define Clear(x,y) memset (x,y,sizeof(x)) #define Close() ios::sync_with_stdio(0) #define Open() freopen ("more.in", "r", stdin) #define get_min(a,b) a = min (a, b) #define get_max(a,b) a = max (a, b); #define y0 yzz #define y1 yzzz #define fi first #define se second #define pii pair<int, int> #define pli pair<long long, int> #define pll pair<long long, long long> #define pb push_back #define pl c<<1 #define pr (c<<1)|1 #define lson tree[c].l,tree[c].mid,pl #define rson tree[c].mid+1,tree[c].r,pr #define mod 1000000007 typedef unsigned long long ull; template <class T> inline T lowbit (T x) {return x&(-x);} template <class T> inline T sqr (T x) {return x*x;} template <class T> inline bool scan (T &ret) { char c; int sgn; if (c = getchar(), c == EOF) return 0; //EOF while (c != '-' && (c < '0' || c > '9') ) c = getchar(); sgn = (c == '-') ? -1 : 1; ret = (c == '-') ? 0 : (c - '0'); while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0'); ret *= sgn; return 1; } const double pi = 3.14159265358979323846264338327950288L; using namespace std; #define INF 1e17 #define maxn 1005 #define maxm 1000005 //-----------------morejarphone--------------------// int n, m; int val[maxn][maxn]; int dp[maxn][maxn][11][11]; void rmq_init () { for(int row = 1; row <= n; row++) for(int col = 1; col <=m; col++) dp[row][col][0][0] = val[row][col]; int mx = log(double(n)) / log(2.0); int my = log(double(m)) / log(2.0); for(int i=0; i<= mx; i++) { for(int j = 0; j<=my; j++) { if(i == 0 && j ==0) continue; for(int row = 1; row+(1<<i)-1 <= n; row++) { for(int col = 1; col+(1<<j)-1 <= m; col++) { if(i == 0)//y轴二分 dp[row][col][i][j]=max(dp[row][col][i][j-1],dp[row][col+(1<<(j-1))][i][j-1]); else//x轴二分 dp[row][col][i][j]=max(dp[row][col][i-1][j],dp[row+(1<<(i-1))][col][i-1][j]); } } } } } int rmq (int x1,int x2,int y1,int y2) { int kx = log(double(x2-x1+1)) / log(2.0); int ky = log(double(y2-y1+1)) / log(2.0); int m1 = dp[x1][y1][kx][ky]; int m2 = dp[x2-(1<<kx)+1][y1][kx][ky]; int m3 = dp[x1][y2-(1<<ky)+1][kx][ky]; int m4 = dp[x2-(1<<kx)+1][y2-(1<<ky)+1][kx][ky]; return max( max(m1,m2) , max(m3,m4)); } int main () { //freopen ("more.in", "r", stdin); scanf ("%d%d", &n, &m); Clear (val, 0); for (int i = 1; i <= n; i++) { for (int j = 1; j <= m; j++) { scan (val[i][j]); } } for (int i = 1; i <= n; i++) { for (int j = 1; j <= m; j++) { if (!val[i][j]) val[i][j] = 0; else val[i][j] = min (min (val[i-1][j], val[i][j-1]), val[i-1][j-1])+1; } } rmq_init (); int q; scan (q); while (q--) { int x1, x2, y1, y2; scanf ("%d%d%d%d", &x1, &y1, &x2, &y2); int l = 0, r = min (x2-x1+1, y2-y1+1); while (r-l > 1) { int mid = (l+r)>>1; if (rmq (x1+mid-1, x2, y1+mid-1, y2) >= mid) l = mid; else r = mid; } if (rmq (x1+r-1, x2, y1+r-1, y2) >= r) printf ("%d\n", r); else printf ("%d\n", l); } return 0; }