題目鏈接:http://poj.org/problem?id=2155
題目大意:
給定n*n的矩陣,q次操作,C操作取反(x1,y1),(x2,y2)爲左上角右下角的子矩陣,Q操作查詢(x,y)單點的值
題目思路:
二維樹狀數組區間修改,單點查詢
AC代碼:
#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]; //行列分開看,每一行每一列都是一個一維樹狀數組
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)){ //列的一維樹狀數組
tree[x][i]+=val;
}
x+=lowbit(x);
}
}
int sum(int x,int y){ //返回(0,0,),(x,y)爲對角頂點的矩陣
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;
}