hdu1828 Picture 扫描线

最新推荐文章于 2021-03-12 11:59:48 发布

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最新推荐文章于 2021-03-12 11:59:48 发布

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分类专栏：数据结构之扫描线

本文链接：https://blog.youkuaiyun.com/wweiainn/article/details/39931957

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本文介绍了HDU1828 Picture问题的解决方案，重点讲解了利用扫描线算法来处理图形的填充。通过实例输入和输出，详细阐述了算法的实现过程，帮助读者理解如何应用扫描线技术解决此类问题。

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Description

A number of rectangular posters, photographs and other pictures of the same shape are pasted on a wall. Their sides are all vertical or horizontal. Each rectangle can be partially or totally covered by the others. The length of the boundary of the union of all rectangles is called the perimeter.

Write a program to calculate the perimeter. An example with 7 rectangles is shown in Figure 1.

The corresponding boundary is the whole set of line segments drawn in Figure 2.

The vertices of all rectangles have integer coordinates.

Input

Your program is to read from standard input. The first line contains the number of rectangles pasted on the wall. In each of the subsequent lines, one can find the integer coordinates of the lower left vertex and the upper right vertex of each rectangle. The values of those coordinates are given as ordered pairs consisting of an x-coordinate followed by a y-coordinate.

0 <= number of rectangles < 5000
All coordinates are in the range [-10000,10000] and any existing rectangle has a positive area.

Output

Your program is to write to standard output. The output must contain a single line with a non-negative integer which corresponds to the perimeter for the input rectangles.

Sample Input

7
-15 0 5 10
-5 8 20 25
15 -4 24 14
0 -6 16 4
2 15 10 22
30 10 36 20
34 0 40 16

Sample Output

做了好几天竟然发现是一个变量写错了。。。

让你秋的是图形的周长，你只需要上下扫描一遍左右扫描一遍就行了。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<math.h>
using namespace std;
int y[100000];
int x[100000];
struct node
{
    int l,r,flag;
    int  yl,yr,s;
} tree[44000];
struct w
{
    int x,yl,yr;
    int flag;
} p[44000],tem,q[44000];
int cmp(struct w a,struct w b)
{
    return a.x<b.x;
}
void build(int left,int right,int root)
{
    tree[root].l=left;
    tree[root].r=right;
    tree[root].s=tree[root].flag=0;
    tree[root].yl=y[left];
    tree[root].yr=y[right];
    if(left+1==right)
        return;
    int mid=(left+right)>>1;
    build(left,mid,root<<1);
    build(mid,right,root<<1|1);
}
void build2(int left,int right,int root)
{
    tree[root].l=left;
    tree[root].r=right;
    tree[root].s=tree[root].flag=0;
    tree[root].yl=x[left];
    tree[root].yr=x[right];
    if(left+1==right)
        return;
    int mid=(left+right)>>1;
    build2(left,mid,root<<1);
    build2(mid,right,root<<1|1);
}
void len(int t)
{
    if(tree[t].flag>0)
    {
        tree[t].s=tree[t].yr-tree[t].yl;
        return ;
    }
    if(tree[t].l+1==tree[t].r)
        tree[t].s=0;
    else
        tree[t].s=tree[t<<1].s+tree[t<<1|1].s;
}
void update(int t,w e)
{
    if(e.yl==tree[t].yl&&e.yr==tree[t].yr)
    {
        tree[t].flag+=e.flag;
        len(t);
        return;
    }
    if(e.yr<=tree[t<<1].yr)
        update(t<<1,e);
    else if(e.yl>=tree[t<<1|1].yl)
        update(t<<1|1,e);
    else
    {
        w tmp=e;
        tmp.yr=tree[t<<1].yr;
        update(t<<1,tmp);
        tmp=e;
        tmp.yl=tree[t<<1|1].yl;
        update(t<<1|1,tmp);
    }
     len(t);
}
int main()
{
    int t,cnt,cnt2;
    int x1,x2,y1,y2;
    while(~scanf("%d",&t)&&t)
    {
        cnt=1;
        cnt2=1;
        while(t--)
        {
            scanf("%d %d %d %d",&x1,&y1,&x2,&y2);
            q[cnt2].x=x1;
            q[cnt2].yl=y1;
            q[cnt2].yr=y2;
            q[cnt2].flag=1;
            x[cnt2++]=y1;
            q[cnt2].x=x2;
            q[cnt2].yl=y1;
            q[cnt2].yr=y2;
            q[cnt2].flag=-1;
            x[cnt2++]=y2;

            p[cnt].x=y1;
            p[cnt].yl=x1;
            p[cnt].yr=x2;
            p[cnt].flag=1;
            y[cnt++]=x1;
            p[cnt].x=y2;
            p[cnt].yl=x1;
            p[cnt].yr=x2;
            p[cnt].flag=-1;
            y[cnt++]=x2;
        }
        sort(p+1,p+cnt,cmp);
        sort(y+1,y+cnt);
        build(1,cnt-1,1);
        update(1,p[1]);
        int  sum=tree[1].s;
        int low=tree[1].s;
        for(int i=2; i<cnt; i++)
        {
            update(1,p[i]);
            sum+=abs(low-tree[1].s);
            low=tree[1].s;
        }
        sort(q+1,q+cnt2,cmp);
        sort(x+1,x+cnt2);
        build2(1,cnt2-1,1);
        update(1,q[1]);
          sum+=tree[1].s;
          low=tree[1].s;
        for(int i=2; i<cnt2; i++)
        {
            update(1,q[i]);
            sum+=abs(low-tree[1].s);
             low=tree[1].s;
        }
        printf("%d\n",sum);
    }
}