Matrix.(POJ-2155)(树状数组)

一道二维树状数组的题目,比较经典,适合新手练习。

可以打印出来每次操作后矩阵的情况,就能很直观的理解这个树状数组是怎么实现的,他将多余的部分巧妙的重复了偶数次,使得多余部分奇偶不会发生变化。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
#include<map>
using namespace std;
int T,n,m,bit[1005][1005];
char s[10];
int sum(int a,int b) {
    int s = 0;
    for(int i=a;i>0;i-=i&-i)
        for(int j=b;j>0;j-=j&-j)
        s+=bit[i][j];
    return s;
}
void add(int a,int b) {
    for(int i=a;i<=n;i+=i&-i)
        for(int j=b;j<=n;j+=j&-j){
            bit[i][j]++;
        }
}
int main() {
    scanf("%d",&T);
    while(T--) {
        scanf("%d%d",&n,&m);
        memset(bit,0,sizeof(bit));
        while(m--) {
            scanf("%s",s);
            if(s[0]=='C'){
                int x1,y1,x2,y2;
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                add(x2+1,y2+1);
                add(x1,y1);
                add(x1,y2+1);
                add(x2+1,y1);
            }
            else {
                int x,y;
                scanf("%d%d",&x,&y);
                printf("%d\n",sum(x,y)%2);
            }
        }
        if(T) printf("\n");
    }
    return 0;
}
时间: 2024-12-06 06:47:21

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