POJ 2155 Matrix (二维线段树)
2014-05-23 23:35
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Matrix
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].
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.
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.
Sample Input
Sample Output
Source
POJ Monthly,Lou Tiancheng
这题二维线段树也可以做。
二维线段树需要 给一个矩形加一个值
查询单个的值。
加值的时候直接加一个块。
查询的时候把这个点以及和这个点相关的都累加起来。
数据结构写法多样啊,重在理解
Time Limit: 3000MS | Memory Limit: 65536K | |
Total Submissions: 17226 | Accepted: 6461 |
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].
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.
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.
Sample Input
1 2 10 C 2 1 2 2 Q 2 2 C 2 1 2 1 Q 1 1 C 1 1 2 1 C 1 2 1 2 C 1 1 2 2 Q 1 1 C 1 1 2 1 Q 2 1
Sample Output
1 0 0 1
Source
POJ Monthly,Lou Tiancheng
这题二维线段树也可以做。
二维线段树需要 给一个矩形加一个值
查询单个的值。
加值的时候直接加一个块。
查询的时候把这个点以及和这个点相关的都累加起来。
数据结构写法多样啊,重在理解
/* *********************************************** Author :kuangbin Created Time :2014/5/23 23:08:19 File Name :E:\2014ACM\专题学习\数据结构\二维线段树\POJ2155.cpp ************************************************ */ #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string> #include <math.h> #include <stdlib.h> #include <time.h> using namespace std; const int MAXN = 1010; struct Nodey { int l,r; int val; }; int n; int locx[MAXN],locy[MAXN]; struct Nodex { int l,r; Nodey sty[MAXN*3]; void build(int i,int _l,int _r) { sty[i].l = _l; sty[i].r = _r; sty[i].val = 0; if(_l == _r) { locy[_l] = i; return; } int mid = (_l + _r)>>1; build(i<<1,_l,mid); build((i<<1)|1,mid+1,_r); } void add(int i,int _l,int _r,int val) { if(sty[i].l == _l && sty[i].r == _r) { sty[i].val += val; return; } int mid = (sty[i].l + sty[i].r)>>1; if(_r <= mid)add(i<<1,_l,_r,val); else if(_l > mid)add((i<<1)|1,_l,_r,val); else { add(i<<1,_l,mid,val); add((i<<1)|1,mid+1,_r,val); } } }stx[MAXN*3]; void build(int i,int l,int r) { stx[i].l = l; stx[i].r = r; stx[i].build(1,1,n); if(l == r) { locx[l] = i; return; } int mid = (l+r)>>1; build(i<<1,l,mid); build((i<<1)|1,mid+1,r); } void add(int i,int x1,int x2,int y1,int y2,int val) { if(stx[i].l == x1 && stx[i].r == x2) { stx[i].add(1,y1,y2,val); return; } int mid = (stx[i].l + stx[i].r)/2; if(x2 <= mid)add(i<<1,x1,x2,y1,y2,val); else if(x1 > mid)add((i<<1)|1,x1,x2,y1,y2,val); else { add(i<<1,x1,mid,y1,y2,val); add((i<<1)|1,mid+1,x2,y1,y2,val); } } int sum(int x,int y) { int ret = 0; for(int i = locx[x];i;i >>= 1) for(int j = locy[y];j;j >>= 1) ret += stx[i].sty[j].val; return ret; } int main() { //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); int T; scanf("%d",&T); while(T--) { int q; scanf("%d%d",&n,&q); build(1,1,n); char op[10]; int x1,x2,y1,y2; while(q--) { scanf("%s",op); if(op[0] == 'C') { scanf("%d%d%d%d",&x1,&y1,&x2,&y2); add(1,x1,x2,y1,y2,1); } else { scanf("%d%d",&x1,&y1); if(sum(x1,y1)%2 == 0)printf("0\n"); else printf("1\n"); } } if(T)printf("\n"); } return 0; }
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