poj 2155 二维线段树或树状数组入门

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Matrix

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 29520 Accepted: 10781

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

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

题意：C:输入一个矩形的左上点和右下点，改变矩形内所有点状态，若是0就改为1，若是1就改为0，Q:查询该点是1还是0

这是个二维树状数组模板题，也是个二维线段树模板题，贴代码

二维树状数组：

#include<stdio.h>#include<string.h>#define maxn 1010int c[maxn][maxn];int n;void update(int x,int y){for(int i=x;i<=n;i+=i&(-i))for(int j=y;j<=n;j+=j&(-j))c[i][j]++;}int query(int x,int y){int ans=0;for(int i=x;i>0;i-=i&(-i))for(int j=y;j>0;j-=j&(-j))ans+=c[i][j];return ans;}int main(){int T;scanf("%d",&T);while(T--){int m,x1,x2,y1,y2,i,j;char ch[2];scanf("%d%d",&n,&m);memset(c,0,sizeof(c));while(m--){scanf("%s",ch);if(ch[0]=='C'){scanf("%d%d%d%d",&x1,&y1,&x2,&y2);update(x1,y1);update(x2+1,y1);update(x1,y2+1);update(x2+1,y2+1);}else{scanf("%d%d",&x1,&y1);int temp=query(x1,y1);if(temp&1)printf("1\n");elseprintf("0\n");}}printf("\n");}}

二维线段树：

#include<stdio.h>#include<string.h>#define maxn 1010int seg[3*maxn][3*maxn];int ans;int n;void query2(int l,int r,int node,int k,int y){ans+=seg[k][node];if(l==r)return;int m=(l+r)/2;if(m>=y)query2(l,m,node*2,k,y);elsequery2(m+1,r,node*2+1,k,y);}void query1(int l,int r,int node,int x,int y){query2(1,n,1,node,y);if(l==r)return;int m=(l+r)/2;if(m>=x)query1(l,m,node*2,x,y);elsequery1(m+1,r,node*2+1,x,y);}void update2(int l,int r,int node,int k,int y1,int y2){if(l>=y1&&r<=y2){seg[k][node]++;return;}int m=(l+r)/2;if(m<y1)update2(m+1,r,node*2+1,k,y1,y2);else if(m>=y2)update2(l,m,node*2,k,y1,y2);else{update2(m+1,r,node*2+1,k,y1,y2);update2(l,m,node*2,k,y1,y2);}}void update1(int l,int r,int node,int x1,int x2,int y1,int y2){if(l>=x1&&r<=x2){update2(1,n,1,node,y1,y2);return;}int m=(l+r)/2;if(m<x1)update1(m+1,r,node*2+1,x1,x2,y1,y2);else if(m>=x2)update1(l,m,node*2,x1,x2,y1,y2);else{update1(m+1,r,node*2+1,x1,x2,y1,y2);update1(l,m,node*2,x1,x2,y1,y2);}}int main(){int T;scanf("%d",&T);while(T--){int m,x1,x2,y1,y2;char ch[2];memset(seg,0,sizeof(seg));scanf("%d%d",&n,&m);while(m--){scanf("%s",ch);if(ch[0]=='Q'){scanf("%d%d",&x1,&y1);ans=0;query1(1,n,1,x1,y1);if(ans&1)printf("1\n");elseprintf("0\n");}else{scanf("%d%d%d%d",&x1,&y1,&x2,&y2);update1(1,n,1,x1,x2,y1,y2);}}printf("\n");}}