首页 > 代码库 > poj 2155 区间更新 单点查询

poj 2155 区间更新 单点查询

2024-07-16 21:25:38   213人阅读

Matrix

Time Limit: 3000 MS Memory Limit: 65536 KB

64-bit integer IO format: %I64d , %I64u Java class name: Main

[Submit] [Status] [Discuss]

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
#include <iostream>#include <stdio.h>#include <string.h>using namespace std;#define max 1005int c[max][max];int n;int lowbit(int x){    return x&-x;}void update(int x,int y,int val){    for(int i=x; i<=n; i+=lowbit(i))    {        for(int j=y; j<=n; j+=lowbit(j))        {            c[i][j]+=val;        }    }}int get_sum(int x,int y){    int s=0;    for(int i=x; i>0; i-=lowbit(i))    {        for(int j=y; j>0; j-=lowbit(j))        {            s+=c[i][j];        }    }    return s;}int main(){   /// freopen("in.txt","r",stdin);    ///freopen("out.txt","w",stdout);   int cc,t,x1,x2,y1,y2,a,b;    char ch;    scanf("%d",&cc);    for(int i=1;i<=cc;i++)    {        memset(c,0,sizeof(c));        scanf("%d%d",&n,&t);        getchar();        for(int j=0;j<t;j++)    ///下标不可以从0开始  会死循环        {            scanf("%c",&ch);            if(ch==‘C‘)            {                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);                getchar();                update(x1,y1,1);                update(x1,y2+1,1);                update(x2+1,y1,1);                update(x2+1,y2+1,1);            }            else            {                scanf("%d%d",&a,&b);                getchar();                printf("%d\n",get_sum(a,b)%2);            }        }        printf("\n");    }    return 0;}

* 矩阵的00在左上角

http://blog.csdn.net/zxy_snow/article/details/6264135   图示


联系
我们