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. #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using nam…

有一个矩阵,每次查询一个子矩阵,判断这个子矩阵的最大值是不是在这个子矩阵的四个角上 裸的二维RMQ #pragma comment(linker, "/STACK:1677721600") #include <map> #include <set> #include <stack> #include <queue> #include <cmath> #include <ctime> #include <vec…

Problem Description Paul draw a big m*n matrix A last month, whose entries Ai,j are all integer numbers ( 1 <= i <= m, 1 <= j <= n ). Now he selects some sub-matrices, hoping to find the maximum number. Then he finds that there may be more tha…

题目链接: Hdu 2888 Check Corners 题目描述: 给出一个n*m的矩阵,问以(r1,c1)为左上角,(r2,c2)为右下角的子矩阵中最大的元素值是否为子矩阵的顶点? 解题思路: 二维区间最值查询,可以用二维的ST算法,dp[x][y][i][j]表示x轴上[x,x+(1<<i)-1]与y轴上[y-(1<<j)+1,y]组成的子矩阵中的最值.预处理的时候处理出来子矩阵的最值,查询的时候对于x,y轴上的查询区间[m, n],都要找到一个k,k满足 n-m+1 <…

<题目链接> <转载于 >>> > 题目大意: 给出一个N*M的矩阵,并且给出该矩阵上每个点对应的值,再进行Q次询问,每次询问给出代询问子矩阵的左上顶点和右下顶点,问该子矩阵的最大值是多少,并且判断该最值是否在该子矩阵的四个顶角上. 解题分析: 很明显求二维区间内的最值,需要用到二维RMQ,其中dp[i][j][k][l]表示左上角为(i,j),右下角为(i + 2 ^ k - 1, j + 2 ^ l - 1)这个矩形内的最值.注意这个四维数组不要开得太大,否则…

Check Corners Time Limit: 2000/10000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2377    Accepted Submission(s): 859 Problem Description Paul draw a big m*n matrix A last month, whose entries Ai,j are all intege…

Check Corners Time Limit: 2000/10000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1646    Accepted Submission(s): 597 Problem Description Paul draw a big m*n matrix A last month, whose entries Ai,j are all integer…

http://acm.hdu.edu.cn/showproblem.php?pid=2888 模板题  直接用二维rmq 读入数据时比较坑爹  cin 会超时 #include <cstdio> #include <cstring> #include <algorithm> using namespace std; ; ][]; void rmq_init(int n, int m) { ; i <= n; i++) { ; j <= m; j++) dp[…

二维RMQ. /* 2888 */ #include <iostream> #include <algorithm> #include <cstdio> #include <cstring> #include <cstdlib> using namespace std; #define MAXN 305 #define MAXM 9 int bit[MAXN]; int dp[MAXN][MAXN][MAXM][MAXM]; int n, m;…

Cornfields Time Limit: 1000MS   Memory Limit: 30000K Total Submissions:8623   Accepted: 4100 Description FJ has decided to grow his own corn hybrid in order to help the cows make the best possible milk. To that end, he's looking to build the cornfiel…