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\n",solve(eh)+solve(ev));
71 return 0;
72 }