POJ2155:Matrix(二维树状数组，经典)「建议收藏」

全栈程序员-站长 • 2022年1月20日下午6:00 • 未分类 • 阅读 37

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

Sample Output

Source

这道题确实非常经典，尤其在这个二进制的计算方面

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

论文里非常具体的证明了

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-站长，转载请注明出处：https://javaforall.net/116489.html原文链接：https://javaforall.net