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!

 #include <iostream>
 #include <memory.h>
 #include <stdio.h>
 #include <algorithm>
 using namespace std;

 class CEdge
 {
 public:
     int y;
     int x1,x2;
     bool isBegin;
     CEdge(int y0=,int x10=,int x20=,bool flag=false):y(y0),x1(x10),x2(x20),isBegin(flag){}
     bool operator <(const CEdge &e2)const
     {
         return y<e2.y || y==e2.y && isBegin;
     }
 };

 int cnt[]={};
 int n;

 int solve(CEdge edges[])
 {
     int ans=;
     memset(cnt,,sizeof cnt);

     sort(edges,edges+*n);
     for(int i=;i<*n;i++)
     {
         CEdge e=edges[i];
         for(int j=e.x1;j<e.x2;j++)
         {
             if(e.isBegin && cnt[j]== || !e.isBegin && cnt[j]==)
                 ans++;

             if(e.isBegin)
                 cnt[j]++;
             else
                 cnt[j]--;
         }
     }
     return ans;
 }

 CEdge eh[],ev[];

 int main()
 {
     freopen("picture.in","r",stdin);
     freopen("picture.out","w",stdout);
     cin>>n;
     for(int i=;i<n;i++)
     {
         int x1,y1,x2,y2;
         scanf("%d %d %d %d",&x1,&y1,&x2,&y2);
         x1+=;
         x2+=;
         y1+=;
         y2+=;

         // 加入数组  横边
         eh[*i]=CEdge(y1,x1,x2,true);
         eh[*i+]=CEdge(y2,x1,x2,false);
         // 加入数组  竖边
         ev[*i]=CEdge(x1,y1,y2,true);
         ev[*i+]=CEdge(x2,y1,y2,false);
     }

     printf("%d\n",solve(eh)+solve(ev));
     return ;
 }