hdoj 1892(二维树状数组)
Problem H
Time Limit : 5000/3000ms (Java/Other) Memory Limit : 65535/32768K (Java/Other)
Total Submission(s) : 8 Accepted Submission(s) : 3
Font: Times New Roman | Verdana | Georgia
Font Size: ← →
Problem Description
When coming into our training room, a lot of books are in my eyes. And every time the books are moving from one place to another one. Now give you the position of the books at the early of the day. And the moving information of the books the day, your work is to tell me how many books are stayed in some rectangles.
To make the problem easier, we divide the room into different grids and a book can only stayed in one grid. The length and the width of the room are less than 1000. I can move one book from one position to another position, take away one book from a position or bring in one book and put it on one position.
Input
For each test case, in the first line there is an Integer Q(1<Q<=100,000), means the queries of the case. Then followed by Q queries.
There are 4 kind of queries, sum, add, delete and move.
For example:
S x1 y1 x2 y2 means you should tell me the total books of the rectangle used (x1,y1)-(x2,y2) as the diagonal, including the two points.
A x1 y1 n1 means I put n1 books on the position (x1,y1)
D x1 y1 n1 means I move away n1 books on the position (x1,y1), if less than n1 books at that position, move away all of them.
M x1 y1 x2 y2 n1 means you move n1 books from (x1,y1) to (x2,y2), if less than n1 books at that position, move away all of them.
Make sure that at first, there is one book on every grid and 0<=x1,y1,x2,y2<=1000,1<=n1<=100.
Output
For each "S" query, just print out the total number of books in that area.
Sample Input
2 3 S 1 1 1 1 A 1 1 2 S 1 1 1 1 3 S 1 1 1 1 A 1 1 2 S 1 1 1 2
Sample Output
Case 1: 1 3 Case 2: 1 4
代码:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int n=1010;
int c[n+1][n+1];
int lowbit(int x)
{
return x&(-x);
}
void update(int x,int y,int val)
{
for(int i=x;i<=n;i+=lowbit(i))
{
for(int j=y;j<=n;j+=lowbit(j))
{
c[i][j]+=val;
}
}
}
int getsum(int x,int y)
{
int cnt=0;
for(int i=x;i>=1;i-=lowbit(i))
{
for(int j=y;j>=1;j-=lowbit(j))
{
cnt+=c[i][j];
}
}
return cnt;
}
int val[n+1][n+1];
int main()
{
int ci;scanf("%d",&ci);
int pl=1;
while(ci--)
{
memset(c,0,sizeof(c));
printf("Case %d:\n",pl++);
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
update(i,j,1);
val[i][j]=1;
}
}
int sa;scanf("%d",&sa);
while(sa--)
{
char ch;cin>>ch;
if(ch=='S')
{
int xx1,yy1,xx2,yy2;
scanf("%d%d%d%d",&xx1,&yy1,&xx2,&yy2);
xx1++,xx2++,yy1++,yy2++;//从1开始
int x1,x2,y1,y2;
x1=min(xx1,xx2);x2=max(xx1,xx2);
y1=min(yy1,yy2);y2=max(yy1,yy2);
int cnt=getsum(x2,y2)-getsum(x1-1,y2)-getsum(x2,y1-1)+getsum(x1-1,y1-1);
printf("%d\n",cnt);
}
else if(ch=='A')
{
int x,y,l;
scanf("%d%d%d",&x,&y,&l);
x++,y++;
update(x,y,l);
val[x][y]+=l;
}
else if(ch=='D')
{
int x,y,l;
scanf("%d%d%d",&x,&y,&l);
x++,y++;
if(l>val[x][y]) l=val[x][y];//important
update(x,y,-l);
val[x][y]+=-l;
}
else
{
int x1,x2,y1,y2,l;
scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&l);
x1++,x2++,y1++,y2++;
if(l>val[x1][y1]) l=val[x1][y1];//important
update(x1,y1,-l);
val[x1][y1]+=-l;
update(x2,y2,l);
val[x2][y2]+=l;
}
}
}
return 0;
}
hdoj 1892(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
- 利用python分析nginx日志
最近在学习python,写了个脚本分析nginx日志,练练手.写得比较粗糙,但基本功能可以实现. 脚本功能:查找出当天访问次数前十位的IP,并获取该IP来源,并将分析结果发送邮件到指定邮箱. 实现前两 ...
- HDU-4570 Multi-bit Trie
http://acm.hdu.edu.cn/showproblem.php?pid=4570 Multi-bit Trie Time Limit: 2000/1000 MS (Java/Others) ...
- iOS不越狱装收费App——注册iOS设备为开发者工具
额,这篇教程主要是我写下来用于总结注册iOS设备和用iResign安装App的过程,想要不越狱安装App当然有办法,但是有几个前提--你是一个Apple开发者,或者你有个朋友是App的开发者.如果没有 ...
- python 3 处理HTTP 请求的包
http http: https://docs.python.org/3/library/http.html http是一个包,里面含有多个模块:http.client,http.server,htt ...
- 364. Nested List Weight Sum II
这个题做了一个多小时,好傻逼. 显而易见计算的话必须知道当前层是第几层,因为要乘权重,想要知道是第几层又必须知道最高是几层.. 用了好久是因为想ONE PASS,尝试过遍历的时候构建STACK,通过和 ...
- 11个让你吃惊的Linux终端命令
- 【转】java 文件 读取目录下的所有文件(包括子目录)
转自:http://www.cnblogs.com/pricks/archive/2009/11/11/1601044.html import java.io.File; import java.io ...
- Linux命令 — 设置或查看网络配置命令ifconfig
ifconfig命令用于设置或查看网络配置,包括IP地址.网络掩码.广播地址等.它是linux系统中,使用频率最高的关于网络方面的命令. 1. 命令介绍 命令格式: ifconfig [interfa ...
- aaalogo写入中文出错的解决方法
一.软件名称: 二.软件用途: 制作小logo 三.问题: aaalog软件不能支持中文输入. 简单的说该软件不能使用中文纯粹是因为字体不支持的原因, 只要导入相应字体就可以 不知道其他人使用aaal ...
- uva 10916 Factstone Benchmark(对数函数的活用)
Factstone Benchmark Amtel has announced that it will release a 128-bit computer chip by 2010, a 256- ...