usaco5.5-Picture

离散化计算重叠矩形的周长。

称平行于x轴的边为横边，我们以横边为例，某一矩形中y坐标比较小的横边我们称为始边，另一边我们称为终边。用一条扫描线从下往上扫描，当扫到一条始边的时候，如果这条始边的正下方出现过k条始边和k条终边，那么这条始边肯定是没被覆盖住的，统计结果;当扫到一条终边的时候，如果这条始边的正下方出现过k条始边和k-1条终边，同理，统计结果。

注意扫描到的边要拆成单位长度的小边分别分析。

Executing...
   Test 1: TEST OK [0.005 secs, 4156 KB]
   Test 2: TEST OK [0.008 secs, 4156 KB]
   Test 3: TEST OK [0.016 secs, 4156 KB]
   Test 4: TEST OK [0.008 secs, 4156 KB]
   Test 5: TEST OK [0.016 secs, 4156 KB]
   Test 6: TEST OK [0.008 secs, 4156 KB]
   Test 7: TEST OK [0.043 secs, 4156 KB]
   Test 8: TEST OK [0.014 secs, 4156 KB]
   Test 9: TEST OK [0.019 secs, 4156 KB]
   Test 10: TEST OK [0.008 secs, 4156 KB]
   Test 11: TEST OK [0.103 secs, 4156 KB]

All tests OK.

Your program ('picture') produced all correct answers! This is your submission #4 for this problem. Congratulations!

 1 #include <iostream>
 2 #include <memory.h>
 3 #include <stdio.h>
 4 #include <algorithm>
 5 using namespace std;
 6 
 7 class CEdge
 8 {
 9 public:
10     int y;
11     int x1,x2;
12     bool isBegin;
13     CEdge(int y0=0,int x10=0,int x20=0,bool flag=false):y(y0),x1(x10),x2(x20),isBegin(flag){}
14     bool operator <(const CEdge &e2)const
15     {
16         return y<e2.y || y==e2.y && isBegin;
17     }
18 };
19 
20 int cnt[40010]={0};
21 int n;
22 
23 int solve(CEdge edges[])
24 {
25     int ans=0;
26     memset(cnt,0,sizeof cnt);
27 
28     sort(edges,edges+2*n);
29     for(int i=0;i<2*n;i++)
30     {
31         CEdge e=edges[i];
32         for(int j=e.x1;j<e.x2;j++)
33         {
34             if(e.isBegin && cnt[j]==0 || !e.isBegin && cnt[j]==1)
35                 ans++;
36 
37             if(e.isBegin)
38                 cnt[j]++;
39             else
40                 cnt[j]--;
41         }
42     }
43     return ans;
44 }
45 
46 CEdge eh[20010],ev[20010];
47 
48 int main()
49 {
50     freopen("picture.in","r",stdin);
51     freopen("picture.out","w",stdout);
52     cin>>n;
53     for(int i=0;i<n;i++)
54     {
55         int x1,y1,x2,y2;
56         scanf("%d %d %d %d",&x1,&y1,&x2,&y2);
57         x1+=10000;
58         x2+=10000;
59         y1+=10000;
60         y2+=10000;
61 
62         // 加入数组  横边
63         eh[2*i]=CEdge(y1,x1,x2,true);
64         eh[2*i+1]=CEdge(y2,x1,x2,false);
65         // 加入数组  竖边
66         ev[2*i]=CEdge(x1,y1,y2,true);
67         ev[2*i+1]=CEdge(x2,y1,y2,false);
68     }
69 
70     printf("%d
",solve(eh)+solve(ev));
71     return 0;
72 }