POJ1195 Mobile phones 【二维线段树】

标签：poj1195

Description

Input

The input is read from standard input as integers and the answers to the queries are written to standard output as integers. The input is encoded as follows. Each input comes on a separate line, and consists of one instruction integer and a number of parameter integers according to the following table. 

The values will always be in range, so there is no need to check them. In particular, if A is negative, it can be assumed that it will not reduce the square value below zero. The indexing starts at 0, e.g. for a table of size 4 * 4, we have 0 <= X <= 3 and 0 <= Y <= 3. 

Table size: 1 * 1 <= S * S <= 1024 * 1024 
Cell value V at any time: 0 <= V <= 32767 
Update amount: -32768 <= A <= 32767 
No of instructions in input: 3 <= U <= 60002 
Maximum number of phones in the whole table: M= 2^30 

Output

Sample Input

0 4
1 1 2 3
2 0 0 2 2 
1 1 1 2
1 1 2 -1
2 1 1 2 3 
3

Sample Output

3
4

附二维树状数组解法：http://blog.csdn.net/chang_mu/article/details/37739053

题意：点更新，区域查询。
题解：基本的二维线段树，（今天看了很久才看懂...线段树扩展到二维果然没有树状数组方便，但是一旦想通，会发现思路其实并不复杂）。

#include <stdio.h>
#define maxn 1026
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1

int tree[maxn * 3][maxn * 3], size;

void getSize(){ scanf("%d", &size); }

void updateY(int rootX, int pos, int val, int l, int r, int rt)
{
	tree[rootX][rt] += val;
	if(l == r) return;
	
	int mid = (l + r) >> 1;
	if(pos <= mid) updateY(rootX, pos, val, lson);
	else updateY(rootX, pos, val, rson);
}

void updateX(int x, int y, int val, int l, int r, int rt)
{
	//更新的任务都交给updateY
	updateY(rt, y, val, 0, size - 1, 1);
	if(l == r) return;
	
	int mid = (l + r) >> 1;
	if(x <= mid) updateX(x, y, val, lson);
	else updateX(x, y, val, rson);
}

void update()
{
	int x, y, val;
	scanf("%d%d%d", &x, &y, &val);
	updateX(x, y, val, 0, size - 1, 1);	
}

int queryY(int rootX, int y1, int y2, int l, int r, int rt)
{
	if(l == y1 && r == y2) return tree[rootX][rt];
	
	int mid = (l + r) >> 1;
	if(y2 <= mid) return queryY(rootX, y1, y2, lson);
	else if(y1 > mid) return queryY(rootX, y1, y2, rson);
	else{
		return queryY(rootX, y1, mid, lson) + queryY(rootX, mid + 1, y2, rson);
	}
}

int queryX(int x1, int x2, int y1, int y2, int l, int r, int rt)
{
	if(x1 == l && x2 == r) return queryY(rt, y1, y2, 0, size - 1, 1);
	
	int mid = (l + r) >> 1;
	if(x2 <= mid) return queryX(x1, x2, y1, y2, lson);
	else if(x1 > mid) queryX(x1, x2, y1, y2, rson);
	else{
		return queryX(x1, mid, y1, y2, lson) + queryX(mid + 1, x2, y1, y2, rson);
	}
}

void query()
{
	int x1, y1, x2, y2;
	scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
	
	printf("%d\n", queryX(x1, x2, y1, y2, 0, size - 1, 1));
}

void (*funArr[])() = {
	getSize, update, query
};

int main()
{	
	int num;
	while(scanf("%d", &num), num != 3)
		(*funArr[num])();
	return 0;
}

POJ1195 Mobile phones 【二维线段树】,布布扣,bubuko.com

POJ1195 Mobile phones 【二维线段树】

标签：poj1195

原文地址：http://blog.csdn.net/chang_mu/article/details/37778895