POJ 2155 Matrix

题目链接：http://poj.org/problem?id=2155

题目大意：

给定n*n的矩阵，q次操作，C操作取反(x1,y1),(x2,y2)为左上角右下角的子矩阵，Q操作查询（x，y）单点的值

题目思路：

二维树状数组区间修改，单点查询

AC代码：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<queue>
#include<stack>
#include<map>

using namespace std;

#define FOU(i,x,y) for(int i=x;i<=y;i++)
#define FOD(i,x,y) for(int i=x;i>=y;i--)
#define MEM(a,val) memset(a,val,sizeof(a))
#define PI acos(-1.0)

const double EXP = 1e-9;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const ll MINF = 0x3f3f3f3f3f3f3f3f;
const double DINF = 0xffffffffffff;
const int mod = 1e9+7;
const int N = 3000;

int tree[N][N];  //行列分开看，每一行每一列都是一个一维树状数组
int n,m;         //n行m列

int lowbit(int x){return x&(-x);}
void add(int x,int y,int val){
    while(x<=n){
        for(int i=y;i<=m;i+=lowbit(i)){  //列的一维树状数组
            tree[x][i]+=val;
        }
        x+=lowbit(x);
    }
}
int sum(int x,int y){  //返回(0,0,),(x,y)为对角顶点的矩阵
    int res=0;
    while(x>0){
        for(int i=y;i>0;i-=lowbit(i)){
            res+=tree[x][i];
        }
        x-=lowbit(x);
    }
    return res;
}
int query(int x1,int y1,int x2,int y2){
    return sum(x2,y2)+sum(x2,y2)-sum(x2,y1-1)-sum(x1-1,y2);  //容斥,注意是否可能超longlong
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    std::ios::sync_with_stdio(false);
    int q,x2,y2,x1,y1,t,val;
    char typ;
    scanf("%d",&t);
    bool flag=false;
    while(t--){
        MEM(tree,0);
        scanf("%d%d",&n,&q);
        m=n;
        while(q--){
            scanf(" %c",&typ);
            if(typ=='C'){
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                add(x1,y1,1);
                add(x2+1,y1,-1);
                add(x1,y2+1,-1);
                add(x2+1,y2+1,1);
            }
            else{
                scanf("%d%d",&x1,&y1);
                printf("%d\n",sum(x1,y1)%2);
            }
        }
        if(flag)
            flag=true;
        else
            puts("");
    }
    return 0;
}