POJ2155:Matrix(二维树状数组，经典)「建议收藏」
POJ2155:Matrix(二维树状数组，经典)「建议收藏」
大家好，又见面了，我是全栈君。
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
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 1Sample Output
1
0
0
1Source
具体的能够參考 《浅谈信息学竞赛中的“0”和“1”》此论文。网上非常多说的并不具体，大多仅仅介绍了翻转。并没有介绍为何sum(x,y)%2能得到结果
http://download.csdn.net/detail/lenleaves/4548401
#include <iostream>#include <stdio.h>#include <string.h>#include <string>#include <stack>#include <queue>#include <map>#include <set>#include <vector>#include <math.h>#include <bitset>#include <list>#include <algorithm>#include <climits>using namespace std;#define lson 2*i#define rson 2*i+1#define LS l,mid,lson#define RS mid+1,r,rson#define UP(i,x,y) for(i=x;i<=y;i++)#define DOWN(i,x,y) for(i=x;i>=y;i--)#define MEM(a,x) memset(a,x,sizeof(a))#define W(a) while(a)#define gcd(a,b) __gcd(a,b)#define LL long long#define N 1005#define INF 0x3f3f3f3f#define EXP 1e-8#define lowbit(x) (x&-x)const int mod = 1e9+7;int c[N][N],n,m,cnt,s,t;int a[N][N];int sum(int x,int y){ int ret = 0; int i,j; for(i = x;i>=1;i-=lowbit(i)) { for(j = y;j>=1;j-=lowbit(j)) { ret+=c[i][j]; } } return ret;}void add(int x,int y){ int i,j; for(i = x;i<=n;i+=lowbit(i)) { for(j = y;j<=n;j+=lowbit(j)) { c[i][j]++; } }}int main(){ int i,j,x,y,ans,t; int x1,x2,y1,y2; char op[10]; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); MEM(c,0); MEM(a,0); while(m--) { scanf("%s",op); if(op[0]=='C') { scanf("%d%d%d%d",&x1,&y1,&x2,&y2); x1++,y1++,x2++,y2++; add(x2,y2); add(x1-1,y1-1); add(x2,y1-1); add(x1-1,y2); } else { scanf("%d%d",&x1,&y1); x2 = x1,y2 = y1; x1++,y1++,x2++,y2++; printf("%d\n",sum(x1,y1)); } } printf("\n"); } return 0;}发布者：全栈程序员栈长，转载请注明出处：https://javaforall.cn/116489.html原文链接：https://javaforall.cn
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