可以说是刚好这个空间大小，卡的刚好。
1 #include <stdio.h> 2 #include <iostream> 3 #include <math.h> 4 #include <algorithm> 5 using namespace std; 6 7 const int N = 302; 8 int n, m; 9 int val[N][N]; 10 int dpmax[N][N][9][9]; 11 //int dpmax[a][b][c][d]; 12 //从val[a][b]-val[a+2^c-1][b+2^d-1];的最大值 13 14 void ST() 15 { 16 int i, j, r, c, k1, k2; 17 for (i = 1; i <= n; i++) 18 for (j = 1; j <= m; j++) 19 dpmax[i][j][0][0] = val[i][j]; 20 k1 = (int)(log(double(n)) / log(2.0)); 21 k2 = (int)(log(double(m)) / log(2.0)); 22 //k1 = log2(n); 23 //k2 = log2(m); 24 for (i = 0; i <= k1; i++) 25 { 26 for (j = 0; j <= k2; j++) 27 { 28 if (i == 0 && j == 0) continue; 29 for (r = 1; r + (1 << i) - 1 <= n; r++) 30 { 31 for (c = 1; c + (1 << j) - 1 <= m; c++) 32 { 33 if (i == 0) 34 dpmax[r][c][i][j] = max(dpmax[r][c][i][j - 1], dpmax[r][c + (1 << (j - 1))][i][j - 1]); 35 else 36 dpmax[r][c][i][j] = max(dpmax[r][c][i - 1][j], dpmax[r + (1 << (i - 1))][c][i - 1][j]); 37 } 38 } 39 } 40 } 41 } 42 43 int query(int r1, int c1, int r2, int c2) 44 { 45 int kr = (int)(log(double(r2 - r1 + 1)) / log(2.0)); 46 int kc = (int)(log(double(c2 - c1 + 1)) / log(2.0)); 47 //k = log2(y-x+1); 48 int t1 = dpmax[r1][c1][kr][kc]; 49 int t2 = dpmax[r2 - (1 << kr) + 1][c1][kr][kc]; 50 int t3 = dpmax[r1][c2 - (1 << kc) + 1][kr][kc]; 51 int t4 = dpmax[r2 - (1 << kr) + 1][c2 - (1 << kc) + 1][kr][kc]; 52 53 return max(max(t1, t2), max(t3, t4)); 54 } 55 56 int main() 57 { 58 int i, j, k; 59 int r1, c1, r2, c2; 60 while (~scanf("%d%d", &n, &m)) 61 { 62 for (i = 1; i <= n; i++) 63 for (j = 1; j <= m; j++) 64 scanf("%d", &val[i][j]); 65 ST();//初始化 66 scanf("%d", &k); 67 while (k--) 68 { 69 scanf("%d%d%d%d", &r1, &c1, &r2, &c2); 70 int ret = query(r1, c1, r2, c2); 71 printf("%d ", ret); 72 if (val[r1][c1] == ret || val[r1][c2] == ret || val[r2][c1] == ret || val[r2][c2] == ret) 73 printf("yes "); 74 else 75 printf("no "); 76 } 77 } 78 return 0; 79 }