POJ - 2155 Matrix (二维树状数组 + 区间修改 + 单点求值 或者 二维线段树 + 区间更新 + 单点求值)

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

1
0
0
1

题目意思是给你一个矩阵，最开始都是为0，然后给你一个左上角的坐标，右下角的坐标，将这个区域的0 -> 1,1 -> 0，然后给你一个点让你求出他此时的结果是1还是0.

这里可以用树状数组的区间修改以及单点求值，只是这里是二维的树状数组，原理是一样的。

将其改变的话，可以如此，假设最开始[1....x2][1....y2]为偶数的话(这里就是看成是前辍和)

先将[1....x2][1....y2]增加1,变为奇数,然后将[1...x1 - 1][y2]增加1变为偶数,将[1....x2][y1]增加1变为偶数,而他们中间重叠的[1....x1 - 1][1....x2 - 1]部分则是被改变了三次，还是奇数，所以还有加1，将他变为偶数,如此除了[x1....x2][y1....y2]变了外，其他的地方依旧没有改变。

/*
Author: 2486
Memory: 4304 KB		Time: 547 MS
Language: G++		Result: Accepted
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 1000 + 5;
int C[MAXN][MAXN];
int N, X, M, x, y, x1, x2, y1, y2;
char op[10];
int lowbits(int x){
return x & (-x);
}

void add(int x,int y){
for(int i = x; i <= N;i += lowbits(i)){
for(int j = y;j <= N;j += lowbits(j)){
C[i][j] ++;
}
}
}

int query(int x,int y){
int ret = 0;
for(int i = x;i > 0;i -= lowbits(i)){
for(int j = y;j > 0;j -= lowbits(j)){
ret += C[i][j];
}
}
return ret;
}

int main(){
//freopen("D://imput.txt","r",stdin);
scanf("%d", &X);
while(X --){
memset(C, 0, sizeof(C));
scanf("%d%d", &N, &M);
while(M --){
scanf("%s", op);
if(op[0] == 'C'){
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
x2 ++;
y2 ++;
add(x1, y1);
add(x1, y2);
add(x2, y1);
add(x2, y2);
}
else{
scanf("%d%d", &x, &y);
printf("%d\n", query(x,y) & 1);//求和即代表求点
}
}
if(X)printf("\n");
}
return 0;
}

线段树代码：

/*Author: 2486Memory: 63688 KB		Time: 1579 MSLanguage: G++		Result: Accepted*/#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#include <queue>using namespace std;#define lson rt << 1, l, mid#define rson rt << 1|1, mid + 1, r#define root 1, 1, Nconst int MAXN = 1000 + 5;int N, T, CASE, x1, y1, x2, y2, ret, x, y;char op[5];int sum[MAXN << 2][MAXN << 2];void update_r(int cr, int y1, int y2, int rt, int l, int r) {if(y1 <= l && r <= y2) {sum[cr][rt] = !sum[cr][rt];return ;}int mid = (l + r) >> 1;if(y1 <= mid) update_r(cr, y1, y2, lson);if(y2 > mid) update_r(cr, y1, y2, rson);}void update_c(int x1, int y1, int x2, int y2, int rt, int l, int r) {if(x1 <= l && r <= x2) {update_r(rt, y1, y2, root);return;}int mid = (l + r) >> 1;if(x1 <= mid) update_c(x1, y1, x2, y2, lson);if(x2 > mid) update_c(x1, y1, x2, y2, rson);}void query_r(int cr, int y1, int y2, int rt, int l, int r) {if(sum[cr][rt]) ret ++;if(y1 <= l && r <= y2) {return ;}int mid = (l + r) >> 1;if(y1 <= mid) query_r(cr, y1, y2, lson);if(y2 > mid) query_r(cr, y1, y2, rson);}void query_c(int x1, int y1, int x2, int y2, int rt, int l, int r) {query_r(rt, y1, y2, root);//选择覆盖了子树的个数if(x1 <= l && r <= x2) {return;}int mid = (l + r) >> 1;if(x1 <= mid) query_c(x1, y1, x2, y2, lson);if(x2 > mid) query_c(x1, y1, x2, y2, rson);}int main() {//freopen("D://imput.txt","r",stdin);scanf("%d", &CASE);while(CASE --) {memset(sum, 0, sizeof(sum));scanf("%d%d", &N, &T);while(T --) {scanf("%s", op);if(op[0] == 'C') {scanf("%d%d%d%d", &x1, &y1, &x2, &y2);update_c(x1, y1, x2, y2, root);} else {scanf("%d%d", &x, &y);ret = 0;query_c(x, y, x, y, root);printf("%d\n", ret & 1);}}if(CASE)printf("\n");}return 0;}