BZOJ 1513 POI 2006 Tet-Tetris 3D 二维线段树

标签：bzoj   poi   二维线段树   数据结构   

题目大意：三维俄罗斯方块，问最后摞了多高。

思路：二维线段树的裸题。但是要注意二维线段树不支持标记下穿。所以就不下传，每次更新答案的时候先看标记，然后用所有的跟标记比较大小之后返回。

具体看代码吧，不知道怎么说。

CODE：

#define _CRT_SECURE_NO_WARNINGS

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 1100
using namespace std;
#define LEFT (pos << 1)
#define RIGHT (pos << 1|1)
int X,Y;

struct SegTreeY{	//数据
	int tree[MAX << 1],tag[MAX << 1];

	int Ask(int l,int r,int x,int y,int pos) {
		if(l == x && r == y)	return tree[pos];
		int mid = (l + r) >> 1,re = tag[pos];
		if(y <= mid)	return max(re,Ask(l,mid,x,y,LEFT));
		if(x > mid)		return max(re,Ask(mid + 1,r,x,y,RIGHT));
		int left = Ask(l,mid,x,mid,LEFT);
		int right = Ask(mid + 1,r,mid + 1,y,RIGHT);
		return max(re,max(left,right));
	}
	void Modify(int l,int r,int x,int y,int c,int pos) {
		tree[pos] = max(tree[pos],c);
		if(l == x && y == r) {
			tag[pos] = max(tag[pos],c);
			return ;
		}
		int mid = (l + r) >> 1;
		if(y <= mid)	Modify(l,mid,x,y,c,LEFT);
		else if(x > mid)	Modify(mid + 1,r,x,y,c,RIGHT);
		else {
			Modify(l,mid,x,mid,c,LEFT);
			Modify(mid + 1,r,mid + 1,y,c,RIGHT);
		}
	}
};

int ask_x,ask_y;

struct SegTreeX{	//寻址
	SegTreeY tree[MAX << 1],tag[MAX << 1];

	int Ask(int l,int r,int x,int y,int pos) {
		if(l == x && y == r)
			return tree[pos].Ask(0,Y,ask_x,ask_y,1);
		int mid = (l + r) >> 1,re = tag[pos].Ask(0,Y,ask_x,ask_y,1);
		if(y <= mid)	return max(re,Ask(l,mid,x,y,LEFT));
		if(x > mid)		return max(re,Ask(mid + 1,r,x,y,RIGHT));
		int left = Ask(l,mid,x,mid,LEFT);
		int right = Ask(mid + 1,r,mid + 1,y,RIGHT);
		return max(re,max(left,right));
	}
	void Modify(int l,int r,int x,int y,int c,int pos) {
		tree[pos].Modify(0,Y,ask_x,ask_y,c,1);
		if(l == x && y == r) {
			tag[pos].Modify(0,Y,ask_x,ask_y,c,1);
			return ;
		}
		int mid = (l + r) >> 1;
		if(y <= mid)	Modify(l,mid,x,y,c,LEFT);
		else if(x > mid)	Modify(mid + 1,r,x,y,c,RIGHT);
		else {
			Modify(l,mid,x,mid,c,LEFT);
			Modify(mid + 1,r,mid + 1,y,c,RIGHT);
		}
	}
}solver;

int cnt;

int main()
{
	cin >> X >> Y >> cnt;
	X += 10,Y += 10;
	for(int d,s,w,x,y,i = 1; i <= cnt; ++i) {
		scanf("%d%d%d%d%d",&d,&s,&w,&x,&y);
		ask_x = y,ask_y = y + s - 1;
		int height = solver.Ask(0,X,x,x + d - 1,1);
		solver.Modify(0,X,x,x + d - 1,height + w,1);
	}
	ask_x = 0,ask_y = Y;
	cout << solver.Ask(0,X,0,X,1) << endl;
	return 0;
}

BZOJ 1513 POI 2006 Tet-Tetris 3D 二维线段树

标签：bzoj   poi   二维线段树   数据结构   

原文地址：http://blog.csdn.net/jiangyuze831/article/details/43964033