HDU 2155 Matrix

Matrix

Time Limit: 3000ms

Memory Limit: 65536KB

This problem will be judged on PKU. Original ID: 2155
64-bit integer IO format: %lld      Java class name: Main

 

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

 

解题：二维线段树，第一次玩这鬼玩意，调试了N久。。。挫。。。。

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = 1001;
struct subtree{
    int lt,rt;
    bool val;
};
struct node{
    int lt,rt;
    subtree sb[maxn<<2];
};
node tree[maxn<<2];
int n,m,ans;
void sub(int lt,int rt,int v,subtree *sb){
    sb[v].lt = lt;
    sb[v].rt = rt;
    sb[v].val = false;
    if(lt == rt) return;
    int mid = (lt + rt)>>1;
    sub(lt,mid,v<<1,sb);
    sub(mid+1,rt,v<<1|1,sb);
}
void build(int lt,int rt,int v){
    tree[v].lt = lt;
    tree[v].rt = rt;
    sub(1,n,1,tree[v].sb);
    if(lt == rt) return;
    int mid = (lt + rt)>>1;
    build(lt,mid,v<<1);
    build(mid+1,rt,v<<1|1);
}
void subupdate(int lt,int rt,int v,subtree *sb){
    if(sb[v].lt >= lt && sb[v].rt <= rt){
        sb[v].val ^= 1;
        return;
    }
    if(lt <= tree[v<<1].rt) subupdate(lt,rt,v<<1,sb);
    if(rt >= tree[v<<1|1].lt) subupdate(lt,rt,v<<1|1,sb);
}
void update(int lt,int rt,int y1,int y2,int v){
    if(tree[v].lt >= lt && tree[v].rt <= rt){
        subupdate(y1,y2,1,tree[v].sb);
        return;
    }
    if(lt <= tree[v<<1].rt) update(lt,rt,y1,y2,v<<1);
    if(rt >= tree[v<<1|1].lt) update(lt,rt,y1,y2,v<<1|1);

}
void subquery(int y,int v,subtree *sb){
    ans ^= sb[v].val;
    if(sb[v].lt == sb[v].rt) return;
    if(y <= sb[v<<1].rt) subquery(y,v<<1,sb);
    else subquery(y,v<<1|1,sb);
}
void query(int x,int y,int v){
    subquery(y,1,tree[v].sb);
    if(tree[v].lt == tree[v].rt) return;
    if(x <= tree[v<<1].rt) query(x,y,v<<1);
    else query(x,y,v<<1|1);
}
int main(){
    int t,x1,y1,x2,y2;
    char s[5];
    scanf("%d",&t);
    while(t--){
        scanf("%d %d",&n,&m);
        build(1,n,1);
        for(int i = 0; i < m; ++i){
            scanf("%s",s);
            if(s[0] == 'Q'){
                ans = 0;
                scanf("%d %d",&x1,&y1);
                query(x1,y1,1);
                printf("%d\n",ans);
            }else if(s[0] == 'C'){
                scanf("%d %d %d %d",&x1,&y1,&x2,&y2);
                update(x1,x2,y1,y2,1);
            }
        }
        puts("");
    }
    return 0;
}