首页 > ACM题库 > HDU-杭电 > HDU 1264 Counting Squares-线段树-[解题报告] C++
2013
12-04

HDU 1264 Counting Squares-线段树-[解题报告] C++

Counting Squares

问题描述 :

Your input is a series of rectangles, one per line. Each rectangle is specified as two points(X,Y) that specify the opposite corners of a rectangle. All coordinates will be integers in the range 0 to 100. For example, the line
5 8 7 10
specifies the rectangle who’s corners are(5,8),(7,8),(7,10),(5,10).
If drawn on graph paper, that rectangle would cover four squares. Your job is to count the number of unit(i.e.,1*1) squares that are covered by any one of the rectangles given as input. Any square covered by more than one rectangle should only be counted once.

输入:

The input format is a series of lines, each containing 4 integers. Four -1′s are used to separate problems, and four -2′s are used to end the last problem. Otherwise, the numbers are the x-ycoordinates of two points that are opposite corners of a rectangle.

输出:

Your output should be the number of squares covered by each set of rectangles. Each number should be printed on a separate line.

样例输入:

5 8 7 10
6 9 7 8
6 8 8 11
-1 -1 -1 -1
0 0 100 100
50 75 12 90
39 42 57 73
-2 -2 -2 -2

样例输出:

8
10000

 一道水的线段树,搞了一个晚上,早上脑子清醒过来后终于找到错的地方了~初始化弱智了,在漆黑的环境里面真的不适合写代码。。。

/*
线段树求矩形面积的并(离散化下先)
把X坐标排序离散,枚举每一个区间,然后扫描线段树,求的覆盖的Y的长度
然后 (X[i+1]-X[i]) *L 即是所求面积
*/
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define NN 300
#define Swap(a,b) (a=a^b,b=b^a,a=a^b)
struct node{
	int r,l,val;
}seg_tree[NN];
struct Link{
	int low_x,low_y,high_x,high_y;
}link[NN];
int X[NN];
int len_l,len_seg,len_x,val,get_ans;
int cmp(const void *a,const void *b){
	return *(int *)a - *(int *)b;
}
void Init(int root,int l,int r)
{
	int mid = (l+r)>>1;
	seg_tree[root].l = l;
	seg_tree[root].r = r;
	seg_tree[root].val = 0;
	len_seg = len_seg < root ? root : len_seg;
	if(r-l == 1)return ;
	Init(root*2,l,mid);
	Init(root*2+1,mid,r);
}
void insert(int root ,int l,int r){
	int mid = (seg_tree[root].l+seg_tree[root].r)>>1;
	if(root > len_seg ) return ;
	if(l <= seg_tree[root].l && r >= seg_tree[root].r)
		seg_tree[root].val = val;
	else if(r<=mid)insert(root*2,l,r);
	else if(l>=mid)insert(root*2+1,l,r);
	else {
		insert(root*2,l,mid);
		insert(root*2+1,mid,r);
	}
	return ;
}
void get_seg(int root)
{
	if(root > len_seg ) return ;
	if(seg_tree[root].val == val){	//用来区分不同区间的Y的长度
		get_ans+=(seg_tree[root].r-seg_tree[root].l);
		return ;
	}
	get_seg(root*2);
	get_seg(root*2+1);
}
int main()
{
	int i,j,ans,x1,x2,y1,y2;	
	Init(1,0,100);
	while(scanf("%d%d%d%d",&x1,&y1,&x2,&y2)){
		ans=len_l=len_x=0;
		if(x1>x2)Swap(x1,x2);
		if(y1>y2)Swap(y1,y2);
		X[len_x++] = x1;
		X[len_x++] = x2;
		link[len_l].low_x = x1;
		link[len_l].low_y = y1;
		link[len_l].high_x = x2;
		link[len_l++].high_y=y2;
		while(scanf("%d%d%d%d",&x1,&y1,&x2,&y2)){
			if(x1 == -1 || x1 == -2)break;
			if(x1>x2)Swap(x1,x2);
			if(y1>y2)Swap(y1,y2);
			X[len_x++] = x1;
			X[len_x++] = x2;
			link[len_l].low_x = x1;
			link[len_l].low_y = y1;
			link[len_l].high_x = x2;
			link[len_l++].high_y=y2;
		}
		qsort(X,len_x,sizeof(int),cmp);
		for(i=0;i<len_x-1;i++){
			if(X[i] == X[i+1])continue;
			val++;//保证了不同区间 线段树中val的值的不同,达到一次建树,多次使用的效果!
			for(j=0;j<len_l;j++){
				if(link[j].low_x <= X[i] && link[j].high_x >= X[i+1])
					insert(1,link[j].low_y,link[j].high_y);
			}
			get_ans=0;
			get_seg(1);
			ans += (X[i+1]-X[i])*get_ans;
		}
		printf("%d\n",ans);
		if(x1 == -2)break;
	}
	return 0;
}

 


  1. 嗯 分析得很到位,确实用模板编程能让面试官对你的印象更好。在设置辅助栈的时候可以这样:push时,比较要push的elem和辅助栈的栈顶,elem<=min.top(),则min.push(elem).否则只要push(elem)就好。在pop的时候,比较stack.top()与min.top(),if(stack.top()<=min.top()),则{stack.pop();min.pop();},否则{stack.pop();}.

  2. “可以发现,树将是满二叉树,”这句话不对吧,构造的树应该是“完全二叉树”,而非“满二叉树”。