1452: [JSOI2009]Count

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BZOJ 1452: [JSOI2009]Count 二维树状数组

1452: [JSOI2009]Count Description Input Output Sample Input Sample Output 1 2 HINT Source 题解:设定C[101][N][N] 树状数组上价值为val的lowbit数组 //meek ///#include<bits/stdc++.h> #include <iostream> #include <cstdio> #include <cmath> #include <…

BZOJ 1452: [JSOI2009]Count(二维BIT)

为每一个权值开一个二维树状数组. ------------------------------------------------------------------------- #include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #define rep(i, n) for(int i = 0; i < n; ++i) #define Rep(i ,n…

二维树状数组 BZOJ 1452 [JSOI2009]Count

题目链接裸二维树状数组 #include <bits/stdc++.h> const int N = 305; struct BIT_2D { int c[105][N][N], n, m; void init(int n, int m) { memset (c, 0, sizeof (c)); this->n = n; this->m = m; } void updata(int k, int x, int y, int z) { for (int i=x; i<=n;…

【BZOJ】1452: [JSOI2009]Count

http://www.lydsy.com/JudgeOnline/problem.php?id=1452 题意:n×m的矩阵上每个点有个颜色,现在有q个操作:1 x y c 将点(x,y)的颜色改为c:2 x1 x2 y1 y2 c 询问矩阵x1y1-x2y2颜色为c的格子数目 #include <bits/stdc++.h> using namespace std; const int N=301; int n, m; int S[101][N][N], col[N][N]; void up…

bzoj 1452: [JSOI2009]Count （二维树状数组）

链接:https://www.lydsy.com/JudgeOnline/problem.php?id=1452 思路: 对每个颜色开一个二维树状数组维护就好了实现代码: #include<bits/stdc++.h> using namespace std; ][][]; ][],n,m,q; int lowbit(int x){ return x&-x; } void add(int x,int y,int z,int rt){ for(int i = x;i <= n;i…

BZOJ 1452 [JSOI2009] Count

这道题好像有点简单的样子... absi找题目好厉害啊...确实是一道比较裸的2dBIT啊. 水掉吧. 附:2dBIT怎么做: 2dBIT就是BIT套BIT啦. 所以修改loop(x+=lowbit(x)){loop(y+=lowbit(y)){}} 查询loop(x-=lowbit(x)){loop(y-=lowbit(y)){}} 然后查询区间当然是用容斥... 假设查询(x1+1,y1+1)(x2,y2) 那么答案=Q(x1,y1)+Q(x2,y2)-Q(x1,y2)-Q(x2,y1) Q…