poj 2155 Matrix(二维线段树,树套树)
来源:互联网 发布:开淘宝店要多少钱2017 编辑:程序博客网 时间:2024/06/03 05:07
Matrix
Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 24131 Accepted: 8930
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 upper-left corner is (x1, y1) and lower-right corner is (x2, y2), we change all the elements in the rectangle by using "not" operation (if it is a '0' then change it into '1' otherwise change it into '0'). To maintain the information of the matrix, you are asked to write a program to receive and execute two kinds of instructions.
1. C x1 y1 x2 y2 (1 <= x1 <= x2 <= n, 1 <= y1 <= y2 <= n) changes the matrix by using the rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2).
2. Q x y (1 <= x, y <= n) querys A[x, y].
We can change the matrix in the following way. Given a rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2), we change all the elements in the rectangle by using "not" operation (if it is a '0' then change it into '1' otherwise change it into '0'). To maintain the information of the matrix, you are asked to write a program to receive and execute two kinds of instructions.
1. C x1 y1 x2 y2 (1 <= x1 <= x2 <= n, 1 <= y1 <= y2 <= n) changes the matrix by using the rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2).
2. Q x y (1 <= x, y <= n) querys A[x, y].
Input
The first line of the input is an integer X (X <= 10) representing the number of test cases. The following X blocks each represents a test case.
The first line of each block contains two numbers N and T (2 <= N <= 1000, 1 <= T <= 50000) representing the size of the matrix and the number of the instructions. The following T lines each represents an instruction having the format "Q x y" or "C x1 y1 x2 y2", which has been described above.
The first line of each block contains two numbers N and T (2 <= N <= 1000, 1 <= T <= 50000) representing the size of the matrix and the number of the instructions. The following T lines each represents an instruction having the format "Q x y" or "C x1 y1 x2 y2", which has been described above.
Output
For each querying output one line, which has an integer representing A[x, y].
There is a blank line between every two continuous test cases.
There is a blank line between every two continuous test cases.
Sample Input
12 10C 2 1 2 2Q 2 2C 2 1 2 1Q 1 1C 1 1 2 1C 1 2 1 2C 1 1 2 2Q 1 1C 1 1 2 1Q 2 1
Sample Output
1001
Source
POJ Monthly,Lou Tiancheng
[Submit] [Go Back] [Status] [Discuss]
题解:二维线段树。
区间修改点查询,因为是异或和,所以我们可以把当前点到跟的路径上的标记累加,就是标记不下放,但是统计时需要计算所有覆盖该点的标记。
#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespace std;int n,m,t,ans;struct data{int treey[4000];int l,r;}treex[4000];void buildy(int x,int now,int l,int r){treex[x].treey[now]=0;if (l==r) return;int mid=(l+r)/2;buildy(x,now<<1,l,mid);buildy(x,now<<1|1,mid+1,r);}void build(int now,int l,int r){buildy(now,1,1,n);if (l==r) return ;int mid=(l+r)/2;build(now<<1,l,mid);build(now<<1|1,mid+1,r);}void update(int x,int now,int l,int r,int ll,int rr){if (l>=ll&&r<=rr){treex[x].treey[now]^=1;return;}int mid=(l+r)/2;if (ll<=mid) update(x,now<<1,l,mid,ll,rr);if (rr>mid) update(x,now<<1|1,mid+1,r,ll,rr);}void change(int now,int l,int r,int ll,int rr,int y1,int y2){if (l>=ll&&r<=rr) { update(now,1,1,n,y1,y2); return;}int mid=(l+r)/2;if (ll<=mid) change(now<<1,l,mid,ll,rr,y1,y2);if (rr>mid) change(now<<1|1,mid+1,r,ll,rr,y1,y2);}void query1(int x,int now,int l,int r,int y){ans^=treex[x].treey[now];if (l==r) return;int mid=(l+r)/2;if (y<=mid) query1(x,now<<1,l,mid,y);else query1(x,now<<1|1,mid+1,r,y);}void query(int now,int l,int r,int x,int y){query1(now,1,1,n,y);if (l==r) return;int mid=(l+r)/2;if (x<=mid) query(now<<1,l,mid,x,y);else query(now<<1|1,mid+1,r,x,y);}int main(){freopen("a.in","r",stdin);scanf("%d",&t);for (int i=1;i<=t;i++){scanf("%d%d",&n,&m);build(1,1,n);for (int j=1;j<=m;j++){char s[10]; scanf("%s",s);if (s[0]=='C'){int x1,y1,x2,y2; scanf("%d%d%d%d",&x1,&y1,&x2,&y2);change(1,1,n,x1,x2,y1,y2);}else{int x,y; scanf("%d%d",&x,&y);ans=0;query(1,1,n,x,y);printf("%d\n",ans);}}printf("\n");}}
0 0
- POJ 2155 Matrix(二维线段树)
- 二维线段树(Matrix,poj 2155)
- POJ 2155 Matrix (二维线段树)
- poj 2155 Matrix(二维线段树,树套树)
- 二维线段树 POJ 2155 Matrix
- POJ 2155 Matrix 二维线段树
- POJ 2155 Matrix【二维线段树】
- POJ 2155 Matrix (二维线段树)
- POJ 2155 Matrix 二维线段树
- POJ 2155 Matrix (二维线段树)
- poj 2155 Matrix 【二维线段树】
- POJ 2155 Matrix (二维线段树)
- POJ 2155 Matrix 二维线段树
- POJ 2155——Matrix(树套树,二维树状数组,二维线段树)
- [poj 2155] Matrix(二维zkw线段树)
- POJ-2155-Matrix(二维树状数组 & 二维线段树)
- POJ-2155:Matrix(二维线段树或二维树状数组)
- POJ 2155 Matrix 二维线段树 区间修改 单点查询
- Material Design(Android6.0)
- 【BZOJ2154】Crash的数字表格,数论练习之二维LCM(莫比乌斯反演)
- HDFS操作
- Material Design设计语言(Android5.X)
- Android之URL
- poj 2155 Matrix(二维线段树,树套树)
- Andrpid VR技术(传感器)
- 机器学习之KNN(K近邻)
- VR技术和Google的Demo
- 教你如何用Unity和Cardboard把3D游戏做成VR游戏
- hdu 4502
- Android中的Apk的加固(加壳)原理解析和实现
- Android SlidingMenu导入ActionBar,Attribute "xxx" has already been defined
- placeholder的内容放到中间