HDU_2642_二维树状数组
Stars
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/65536 K (Java/Others)
Total Submission(s): 1628 Accepted Submission(s):
683
stars in the sky.
To make the problem easier,we considerate the sky is a
two-dimension plane.Sometimes the star will be bright and sometimes the star
will be dim.At first,there is no bright star in the sky,then some information
will be given as "B x y" where 'B' represent bright and x represent the X
coordinate and y represent the Y coordinate means the star at (x,y) is
bright,And the 'D' in "D x y" mean the star at(x,y) is dim.When get a query as
"Q X1 X2 Y1 Y2",you should tell Yifenfei how many bright stars there are in the
region correspond X1,X2,Y1,Y2.
There is only one case.
followed.
each line start with a operational character.
if the character
is B or D,then two integer X,Y (0 <=X,Y<= 1000)followed.
if the
character is Q then four integer X1,X2,Y1,Y2(0 <=X1,X2,Y1,Y2<= 1000)
followed.
line.
void add(int k,int x)
{
for(int i=k;i<MAXN;i+=lowbit(i))
c[i]+=x;
}
lowbit():
int lowbit(int x) //取最低位
{
return x&(-x);
}
查询:
int get_sum(int k)
{
int res=;
for(int i=k;i>;i-=lowbit(i))
res+=c[i];
return res;
}
这道题是一道二维树状数组,原理其实也就是这样。
注意:题目中坐标从0开始,可能对一颗star做两次同样的操作。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<algorithm>
#include<stdlib.h>
#include<stack>
#include<vector>
using namespace std; const int MAXN=;
int a[MAXN][MAXN];
bool b[MAXN][MAXN]; int lowbit(int x)
{
return x&(-x);
} void modify(int x,int y,int data)
{
for(int i=x; i<MAXN; i+=lowbit(i))
for(int j=y; j<MAXN; j+=lowbit(j))
a[i][j]+=data;
} int getsum(int x,int y)
{
int res=;
for(int i=x; i>; i-=lowbit(i))
for(int j=y; j>; j-=lowbit(j))
res+=a[i][j];
return res;
} int main()
{
int n,x,y,x1,y1;
char str[];
memset(a,,sizeof(a));
memset(b,,sizeof(b));
scanf("%d",&n);
while(n--)
{
scanf("%s",str);
if(str[]=='B')
{
scanf("%d%d",&x,&y);
x++;
y++;
if(b[x][y]) continue;
modify(x,y,);
b[x][y]=;
}
else if(str[]=='D')
{
scanf("%d%d",&x,&y);
x++;
y++;
if(b[x][y]==) continue;
modify(x,y,-);
b[x][y]=;
}
else
{
scanf("%d%d%d%d",&x,&x1,&y,&y1);
x++;x1++;y++;y1++;
if(x>x1) swap(x,x1);
if(y>y1) swap(y,y1);
int ans=getsum(x1,y1)-getsum(x-,y1)-getsum(x1,y-)+getsum(x-,y-);
printf("%d\n",ans);
}
}
return ;
}
HDU_2642_二维树状数组的更多相关文章
- 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, ...
- HDU1559 最大子矩阵 (二维树状数组)
题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1559 最大子矩阵 Time Limit: 30000/10000 MS (Java/Others) ...
- POJMatrix(二维树状数组)
Matrix Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 22058 Accepted: 8219 Descripti ...
- poj 1195:Mobile phones(二维树状数组,矩阵求和)
Mobile phones Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 14489 Accepted: 6735 De ...
- Codeforces Round #198 (Div. 1) D. Iahub and Xors 二维树状数组*
D. Iahub and Xors Iahub does not like background stories, so he'll tell you exactly what this prob ...
- POJ 2155 Matrix(二维树状数组+区间更新单点求和)
题意:给你一个n*n的全0矩阵,每次有两个操作: C x1 y1 x2 y2:将(x1,y1)到(x2,y2)的矩阵全部值求反 Q x y:求出(x,y)位置的值 树状数组标准是求单点更新区间求和,但 ...
- [poj2155]Matrix(二维树状数组)
Matrix Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 25004 Accepted: 9261 Descripti ...
- POJ 2155 Matrix (二维树状数组)
Matrix Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 17224 Accepted: 6460 Descripti ...
- [POJ2155]Matrix(二维树状数组)
题目:http://poj.org/problem?id=2155 中文题意: 给你一个初始全部为0的n*n矩阵,有如下操作 1.C x1 y1 x2 y2 把矩形(x1,y1,x2,y2)上的数全部 ...
随机推荐
- [bzoj2850]巧克力王国_KD-Tree
巧克力王国 bzoj-2850 题目大意:给出n块巧克力,每块巧克力都有自己的两个参数x和y和本身的价值val,询问:m个人,每个人有两个系数和一个限度a,b,和c.求所有ax+by<=c的巧克 ...
- ASP.NET MVC 源码分析(二) —— 从 IRouteBuilder认识路由构建
我们来看IRouteBuilder的定义: public interface IRouteBuilder { IRouter DefaultHandler { get; set; } IService ...
- Maven中的dependency的scope作用域详解
1.test范围指的是测试范围有效,在编译和打包时都不会使用这个依赖 2.compile范围指的是编译范围有效,在编译和打包时都会将依赖存储进去 3.provided依赖:在编译和测试的过程有效,最后 ...
- oracle 学习笔记之触发器
说明 数据库触发器是一个与表相关联的.存储的PL/SQL程序. 每当一个特定的数据操作语句(Insert,update,delete)在指定的表上发出时,Oracle自己主动地运行触发器中定义的语句序 ...
- [深入学习C#]输入输出安全性——可变类型形參列表的变化安全性
可变类型形參列表(variant-type-parameter-lists) 可变类型形參列表(variant-type-parameter-lists )仅仅能在接口和托付类型上出现.它与普通的ty ...
- edittext禁止android软键盘弹出
1. EditText ed=(EditText) findViewById(R.id.test); ed.clearFocus(); 2. 在AndroidMainfest.xml中选择哪个acti ...
- [APIO2008]DNA
https://zybuluo.com/ysner/note/1158123 题面 戳我 解析 我们要求出第\(r\)种方案,莫过于看其前面什么时候有\(r-1\)种方案. 于是,我们要求出每种情况的 ...
- IJ:ALI OSS 配置
ylbtech-IJ:ALI OSS 配置 1. src/resources/返回顶部 1.src/resources/ 1.1.aliyunoss.properties # oss\u7684\u5 ...
- Java 解析Json数据
Json格式字符串{success:0,errorMsg:"错误消息",data:{total:"总记录数",rows:[{id:"任务ID" ...
- 原生JS---5
原生js学习笔记5——BOM操作 什么是BOM BOM:Browser Object Model 是浏览器对象模型,浏览器对象模型提供了独立与内容的.可以与浏览器窗口进行互动的对象结构,BOM由多个对 ...