HDU_2888_Check Corners】的更多相关文章

Check Corners Time Limit: 2000/10000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3247    Accepted Submission(s): 1173 Problem Description Paul draw a big m*n matrix A last month, whose entries Ai,j are all intege…
CSS3 Rounded Corners The border-radius property is a shorthand property for setting the four border-*-radius properties. syntax border-radius: 1-4 length|% / 1-4 length|%|initial|inherit; Property Description border-radius A shorthand property for se…
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 than one ma…
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…
Check Corners Time Limit: 2000/10000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 858    Accepted Submission(s): 275 Problem Description Paul draw a big m*n matrix A last month, whose entries Ai,j are all integer…
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 <…
网上查找资料 记录学习 int strokeWidth = 5; // 3dp 边框宽度 int roundRadius = 15; // 8dp 圆角半径 int strokeColor = Color.parseColor("#2E3135");//边框颜色 int fillColor = Color.parseColor("#DFDFE0");//内部填充颜色 GradientDrawable gd = new GradientDrawable();//创建d…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2888 模板题.解题思路如下(转载别人写的): dp[row][col][i][j] 表示[row,row+2^i-1]x[col,col+2^j-1] 二维区间内的最小值这是RMQ-ST算法的核心: 倍增思想== min( [row,row+ 2^(i-1)-1]x[col,col+2^j-1], [row+2^(i-1),row+2^i-1]x[col,col+2^j-1] )= min(dp[r…
1.Border radius The border-radius property can be used to working clockwise from top-left set border-top-left-radius/border-top-right-radius/border-bottom-right-radius/border-bottom-left-radius -webkit-border-radius and -moz-border-radius ,old browse…
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;…