• POJ2155 Matrix


     
    Time Limit: 3000MS   Memory Limit: 65536K
    Total Submissions: 25080   Accepted: 9293

    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 1
    

    Sample Output

    1
    0
    0
    1
    

    Source

    POJ Monthly,Lou Tiancheng
     
    二维树状数组果然比二维线段树简单多了
    讲一下二维树状数组
    其实我也不清楚多出来的一维怎么做
    但既然多套了一重循环就算作是二维了
    文字说不清,自己仿照一维画一个图就明白了
    再说这道题
    假设只有一维,我们可以用树状数组维护一个差分数组,区间首尾打标记+1,求和即可
    那么推广到二维,把维护差分数组的方式看成打一个标记
    四个点+1,对询问求一遍和模2
    尹神的办法zrl说可以推广,而这种办法只对01有效
    就是说对于正常的差分数组,区间修改应该是首加尾减
    到了二维应该这样维护
    -1 +1
    +1 -1
    就这样吧
     
    ——by隔壁老司机RLQ
     
     
     1 /*by SilverN*/
     2 #include<algorithm>
     3 #include<iostream>
     4 #include<cstring>
     5 #include<cstdio>
     6 #include<cmath>
     7 using namespace std;
     8 const int mxn=1200;
     9 int t[mxn][mxn];
    10 int n,m;
    11 int lowbit(int x){return x&-x;}
    12 int an;
    13 void add(int x,int y,int v){
    14     while(x<=n){
    15         int tmp=y;
    16         while(tmp<=n){ t[x][tmp]+=v; tmp+=lowbit(tmp);}
    17         x+=lowbit(x);
    18     }
    19 }
    20 int sum(int x,int y){
    21     int res=0;
    22     while(x){
    23         int tmp=y;
    24         while(tmp){    res+=t[x][tmp];     tmp-=lowbit(tmp);}
    25         x-=lowbit(x);
    26     }
    27     return res;
    28 }
    29 int ask(int x1,int y1,int x2,int y2){
    30     return sum(x2,y2)+sum(x1-1,y1-1)-sum(x2,y1-1)-sum(x1,y2-1);
    31 }
    32 int main(){
    33     int T;
    34     scanf("%d",&T);
    35     int i,j;
    36     char ch[5];
    37     int x1,y1,x2,y2;
    38     while(T--){
    39         memset(t,0,sizeof t);
    40         scanf("%d%d",&n,&m);
    41         an=n+5;
    42         for(i=1;i<=m;i++){
    43             scanf("%s",ch);
    44             if(ch[0]=='C'){
    45                 scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
    46                 add(x1,y1,1);
    47                 add(x1,y2+1,-1);
    48                 add(x2+1,y1,-1);
    49                 add(x2+1,y2+1,1);
    50             }
    51             else{
    52                 scanf("%d%d",&x1,&y1);
    53                 printf("%d
    ",sum(x1,y1)%2);
    54             }
    55         }
    56         printf("
    ");
    57     }
    58     return 0;
    59 }
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  • 原文地址:https://www.cnblogs.com/SilverNebula/p/5873156.html
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