Matrix Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 20599   Accepted: 7673 Description Given an N*N matrix A, whose elements are either 0 or 1. A[i, j] means the number in the i-th row and j-th column. Initially we have A[i, j] = 0 (1…

题意:给你一个矩阵开始全是0,然后给你两种指令,第一种:C x1,y1,x2,y2 就是将左上角为x1,y1,右下角为x2,y2,的这个矩阵内的数字全部翻转,0变1,1变0 第二种:Q x1 y1,输出举证x1,y1,位置的数值 思路:该题的巧妙之处是反转的重叠抵消,记住要向上统计,向下修改:二维情况,最主要的是理解那个数组中的每个点保存的值的意义(一个矩形区域的总和):最后记住取余: 此题参考博客 http://blog.csdn.net/xymscau/article/details/660…

题目链接:http://poj.org/problem?id=2155 Matrix Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 32950   Accepted: 11943 Description Given an N*N matrix A, whose elements are either 0 or 1. A[i, j] means the number in the i-th row and j-th col…

Matrix Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 21757   Accepted: 8141 Description Given an N*N matrix A, whose elements are either 0 or 1. A[i, j] means the number in the i-th row and j-th column. Initially we have A[i, j] = 0 (1…

题意: 有一个n*n的矩阵,初始化全部为0.有2中操作: 1.给一个子矩阵,将这个子矩阵里面所有的0变成1,1变成0:2.询问某点的值 方法一:二维线段树 参考链接: http://blog.csdn.net/xiamiwage/article/details/8030273 思路: 二维线段树,一维线段树的成段更新需要lazy. 引申到二维线段树应该需要一个lazy,一个sublazy,可是这里什么都不用.    奇妙之处在于这题的操作是异或,当某一段区间需要异或操作时候, 不必更新到它所有的…

Matrix Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 25004   Accepted: 9261 Description Given an N*N matrix A, whose elements are either 0 or 1. A[i, j] means the number in the i-th row and j-th column. Initially we have A[i, j] = 0 (1…

Description Given an N*N matrix A, whose elements are either 0 or 1. A[i, j] means the number in the i-th row and j-th column. Initially we have A[i, j] = 0 (1 <= i, j <= N). We can change the matrix in the following way. Given a rectangle whose upp…

思路: 大力出奇迹,先用二维树状数组存,然后暴力枚举 算某个矩形区域的值的示意图如下,代码在下面慢慢找... 代码: #include<cstdio> #include<map> #include<cstring> #include<algorithm> #define ll long long using namespace std; const int N = 1030+5; int bit[N][N],n,m; int lowbit(int x){ r…

Matrix Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 16950   Accepted: 6369 Description Given an N*N matrix A, whose elements are either 0 or 1. A[i, j] means the number in the i-th row and j-th column. Initially we have A[i, j] = 0 (1…

这题反反复复,到现在才过. 这道题就是树状数组的逆用,用于修改区间内容,查询点的值. 如果单纯就这个奇偶数来判的话,似乎这个思路比较好理解. 看了一下国家集训队论文(囧),<关于0与1在信息学奥赛中的运用>,. 还有这题卡在输入输出好久. update(a,b,1); update(a,d,-1); update(c,b,-1); update(c,d,1);  用来判二维的情况!! #include <iostream> #include <stdio.h> #inc…