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【HDU 1828】 Picture （矩阵周长并，线段树，扫描法）

【题目】

Picture

Problem 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.

Please process to the end of file.

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

228

Source

IOI 1998

【题意】

　　看图就懂，求矩形周长并。

【分析】

　　很容易想到x和y分开搞，其实就是线段的并咯，但是要注意他有退出的一刻。

　　进去和退出都有可能影响答案，但是不影响答案就不需要加进去ans。

  1 #include<cstdio>
  2 #include<cstdlib>
  3 #include<cstring>
  4 #include<iostream>
  5 #include<algorithm>
  6 #include<cmath>
  7 using namespace std;
  8 #define Maxn 100010
  9 
 10 struct hp
 11 {
 12     int x1,x2,flag;
 13     double y;
 14 }t[Maxn*2];int tl;
 15 struct lsh
 16 {
 17     double x;
 18     int id;
 19 }q[Maxn*2];int ql;
 20 
 21 bool cmp(lsh x,lsh y) {return x.x<y.x;} 
 22 bool cmp2(hp x,hp y) {return (x.y==y.y)?(x.flag>y.flag):(x.y<y.y);}
 23 
 24 int mymin(int x,int y) {return x<y?x:y;}
 25 
 26 double dis[2*Maxn];
 27 
 28 struct node
 29 {
 30     int l,r,lc,rc,cnt;
 31     double sm;
 32 }tr[Maxn*2];int len;
 33 int build(int l,int r)
 34 {
 35     int x=++len;
 36     tr[x].l=l;tr[x].r=r;
 37     tr[x].sm=tr[x].cnt=0;
 38     if(l<r-1)
 39     {
 40         int mid=(tr[x].l+tr[x].r)>>1;
 41         tr[x].lc=build(l,mid);
 42         tr[x].rc=build(mid,r);
 43     }
 44     else tr[x].lc=tr[x].rc=0;
 45     return x;
 46 }
 47 
 48 void upd(int x)
 49 {
 50     int l=tr[x].l,r=tr[x].r,lc=tr[x].lc,rc=tr[x].rc;
 51     if(tr[x].cnt>=1) tr[x].sm=dis[r]-dis[l];
 52     else tr[x].sm=tr[lc].sm+tr[rc].sm;
 53 }
 54 
 55 void change(int x,int l,int r,int c)
 56 {
 57     if(tr[x].l==l&&tr[x].r==r)
 58     {
 59         tr[x].cnt+=c;
 60         upd(x);
 61         return;
 62     }
 63     int mid=(tr[x].l+tr[x].r)>>1;
 64     if(r<=mid) change(tr[x].lc,l,r,c);
 65     else if(l>=mid) change(tr[x].rc,l,r,c);
 66     else
 67     {
 68         change(tr[x].lc,l,mid,c);
 69         change(tr[x].rc,mid,r,c);
 70     }
 71     upd(x);
 72 }
 73 
 74 double x1[Maxn],ya[Maxn],x2[Maxn],y2[Maxn];
 75 int n;
 76 void init()
 77 {
 78     for(int i=1;i<=n;i++) scanf("%lf%lf%lf%lf",&x1[i],&ya[i],&x2[i],&y2[i]);
 79 }
 80 
 81 double ans;
 82 
 83 void ffind()
 84 {
 85     tl=0;ql=0;
 86     for(int i=1;i<=n;i++)
 87     {
 88         //t[++tl].x1,
 89         t[++tl].flag=1;t[tl].y=ya[i];
 90         t[++tl].flag=-1;t[tl].y=y2[i];
 91         q[++ql].x=x1[i],q[ql].id=tl;
 92         q[++ql].x=x2[i],q[ql].id=tl+2*n;
 93     }
 94     sort(q+1,q+1+ql,cmp);
 95     int p=0;
 96     for(int i=1;i<=ql;i++)
 97     {
 98         if(q[i].x!=q[i-1].x||p==0) p++,dis[p]=dis[p-1]+q[i].x-q[i-1].x;
 99         if(q[i].id<=2*n) t[q[i].id].x1=t[q[i].id-1].x1=p;
100         else t[q[i].id-2*n].x2=t[q[i].id-1-2*n].x2=p;
101     }
102     sort(t+1,t+1+tl,cmp2);
103     len=0;tr[0].sm=0;
104     build(1,p);
105     double now=0;
106     for(int i=1;i<=tl;i++)
107     {
108         change(1,t[i].x1,t[i].x2,t[i].flag);
109         double nn=tr[1].sm;
110         if(t[i].flag==1) ans+=nn-now;
111         else ans+=now-nn;
112         now=nn;
113     }
114 }
115 
116 int main()
117 {
118     int T;
119     // scanf("%d",&T);
120     // T=1;
121     while(scanf("%d",&n)!=EOF)
122     {
123         
124         init();
125         ans=0;
126         ffind();
127         // printf("%.0lf
",ans);
128         for(int i=1;i<=n;i++)
129         {
130             double tt;
131             tt=x1[i],x1[i]=ya[i],ya[i]=tt;
132             tt=x2[i],x2[i]=y2[i],y2[i]=tt;
133         }
134         ffind();
135         printf("%.0lf
",ans);
136     }
137     return 0;
138 }

View Code

ws的题目，谁说了多组啊！！！

哦，对了，有一个比较坑的地方【ORZ。。GDXB

矩形边界重合的时候的处理。举个例子，在处理x轴的时候：

中间那条一次也不算。

这个东西，在排序的时候注意一下就好了。

  1 #include<cstdio>
  2 #include<cstdlib>
  3 #include<cstring>
  4 #include<iostream>
  5 #include<algorithm>
  6 #include<cmath>
  7 using namespace std;
  8 #define Maxn 100010
  9 
 10 struct hp
 11 {
 12     int x1,x2,flag;
 13     double y;
 14 }t[Maxn*2];int tl;
 15 struct lsh
 16 {
 17     double x;
 18     int id;
 19 }q[Maxn*2];int ql;
 20 
 21 bool cmp(lsh x,lsh y) {return x.x<y.x;} 
 22 bool cmp2(hp x,hp y) {return (x.y==y.y)?(x.flag>y.flag):(x.y<y.y);}
 23 
 24 int mymin(int x,int y) {return x<y?x:y;}
 25 
 26 double dis[2*Maxn];
 27 
 28 struct node
 29 {
 30     int l,r,lc,rc,cnt;
 31     double sm;
 32 }tr[Maxn*2];int len;
 33 int build(int l,int r)
 34 {
 35     int x=++len;
 36     tr[x].l=l;tr[x].r=r;
 37     tr[x].sm=tr[x].cnt=0;
 38     if(l<r-1)
 39     {
 40         int mid=(tr[x].l+tr[x].r)>>1;
 41         tr[x].lc=build(l,mid);
 42         tr[x].rc=build(mid,r);
 43     }
 44     else tr[x].lc=tr[x].rc=0;
 45     return x;
 46 }
 47 
 48 void upd(int x)
 49 {
 50     int l=tr[x].l,r=tr[x].r,lc=tr[x].lc,rc=tr[x].rc;
 51     if(tr[x].cnt>=1) tr[x].sm=dis[r]-dis[l];
 52     else tr[x].sm=tr[lc].sm+tr[rc].sm;
 53 }
 54 
 55 void change(int x,int l,int r,int c)
 56 {
 57     if(tr[x].l==l&&tr[x].r==r)
 58     {
 59         tr[x].cnt+=c;
 60         upd(x);
 61         return;
 62     }
 63     int mid=(tr[x].l+tr[x].r)>>1;
 64     if(r<=mid) change(tr[x].lc,l,r,c);
 65     else if(l>=mid) change(tr[x].rc,l,r,c);
 66     else
 67     {
 68         change(tr[x].lc,l,mid,c);
 69         change(tr[x].rc,mid,r,c);
 70     }
 71     upd(x);
 72 }
 73 
 74 double x1[Maxn],ya[Maxn],x2[Maxn],y2[Maxn];
 75 int n;
 76 void init()
 77 {
 78     for(int i=1;i<=n;i++) scanf("%lf%lf%lf%lf",&x1[i],&ya[i],&x2[i],&y2[i]);
 79 }
 80 
 81 double ans;
 82 
 83 void ffind()
 84 {
 85     tl=0;ql=0;
 86     for(int i=1;i<=n;i++)
 87     {
 88         //t[++tl].x1,
 89         t[++tl].flag=1;t[tl].y=ya[i];
 90         t[++tl].flag=-1;t[tl].y=y2[i];
 91         q[++ql].x=x1[i],q[ql].id=tl;
 92         q[++ql].x=x2[i],q[ql].id=tl+2*n;
 93     }
 94     sort(q+1,q+1+ql,cmp);
 95     int p=0;
 96     for(int i=1;i<=ql;i++)
 97     {
 98         if(q[i].x!=q[i-1].x||p==0) p++,dis[p]=dis[p-1]+q[i].x-q[i-1].x;
 99         if(q[i].id<=2*n) t[q[i].id].x1=t[q[i].id-1].x1=p;
100         else t[q[i].id-2*n].x2=t[q[i].id-1-2*n].x2=p;
101     }
102     sort(t+1,t+1+tl,cmp2);
103     len=0;tr[0].sm=0;
104     build(1,p);
105     double now=0;
106     for(int i=1;i<=tl;i++)
107     {
108         change(1,t[i].x1,t[i].x2,t[i].flag);
109         double nn=tr[1].sm;
110         if(t[i].flag==1) ans+=nn-now;
111         else ans+=now-nn;
112         now=nn;
113     }
114 }
115 
116 int main()
117 {
118     int T;
119     // scanf("%d",&T);
120     // T=1;
121     while(scanf("%d",&n)!=EOF)
122     {
123         
124         init();
125         ans=0;
126         ffind();
127         // printf("%.0lf
",ans);
128         for(int i=1;i<=n;i++)
129         {
130             double tt;
131             tt=x1[i],x1[i]=ya[i],ya[i]=tt;
132             tt=x2[i],x2[i]=y2[i],y2[i]=tt;
133         }
134         ffind();
135         printf("%.0lf
",ans);
136     }
137     return 0;
138 }

View Code

2016-11-10 15:49:48

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原文地址：https://www.cnblogs.com/Konjakmoyu/p/6051216.html