标签:ring 最大值 模板 ble target problem pre ret 没有
http://acm.hdu.edu.cn/showproblem.php?pid=2888
题意:给出一个n*m的矩阵,还有q个询问,对于每个询问有一对(x1,y1)和(x2,y2),求这个子矩阵中的最大值,和判断四个角有没有等于这个最大值的。
思路:二维RMQ模板题。注意内存卡的挺紧的。
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <cmath> 5 using namespace std; 6 #define N 301 7 int dp[N][N][9][9], mp[N][N], n, m, x1, x2, y11, y2; 8 // 内存刚好 30704kB 卡着过了 9 10 void Init() { 11 for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) dp[i][j][0][0] = mp[i][j]; 12 int A = (int)(log(n) / log(2)), B = (int)(log(m) / log(2)); 13 for(int j = 0; j <= A; j++) { 14 for(int y = 0; y <= B; y++) { 15 if(j == 0 && y == 0) continue; 16 for(int i = 1; i + (1 << j) - 1 <= n; i++) 17 for(int x = 1; x + (1 << y) - 1 <= m; x++) 18 if(j) dp[i][x][j][y] = max(dp[i][x][j-1][y], dp[i+(1<<(j-1))][x][j-1][y]); 19 else dp[i][x][j][y] = max(dp[i][x][j][y-1], dp[i][x+(1<<(y-1))][j][y-1]); 20 } 21 } 22 } 23 24 int Query() { 25 int A = (int)(log(x2 - x1 + 1) / log(2)), B = (int)(log(y2 - y11 + 1) / log(2)); 26 int a = dp[x1][y11][A][B]; 27 int b = dp[x2-(1<<A)+1][y11][A][B]; 28 int c = dp[x1][y2-(1<<B)+1][A][B]; 29 int d = dp[x2-(1<<A)+1][y2-(1<<B)+1][A][B]; 30 return max(a, max(b, max(c, d))); 31 } 32 33 bool check(int val) { 34 if(mp[x1][y11] == val || mp[x2][y2] == val || mp[x1][y2] == val || mp[x2][y11] == val) return true; 35 return false; 36 } 37 38 int main() { 39 while(~scanf("%d%d", &n, &m)) { 40 for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) scanf("%d", &mp[i][j]); 41 int q; scanf("%d", &q); 42 Init(); 43 while(q--) { 44 scanf("%d%d%d%d", &x1, &y11, &x2, &y2); 45 int val = Query(); 46 printf("%d ", val); 47 if(check(val)) puts("yes"); 48 else puts("no"); 49 } 50 } 51 return 0; 52 }
标签:ring 最大值 模板 ble target problem pre ret 没有
原文地址:http://www.cnblogs.com/fightfordream/p/6344862.html