POJ1177 Picture —— 求矩形并的周长 线段树 + 扫描线 + 离散化

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <cmath>
  5 #include <algorithm>
  6 #include <vector>
  7 #include <queue>
  8 #include <stack>
  9 #include <map>
 10 #include <string>
 11 #include <set>
 12 using namespace std;
 13 typedef long long LL;
 14 const double EPS = 1e-8;
 15 const int INF = 2e9;
 16 const LL LNF = 2e18;
 17 const int MAXN = 1e4+10;
 18 
 19 struct line
 20 {
 21     int le, ri, h;
 22     int id;
 23     bool operator<(const line &a)const{
 24         return h<a.h;
 25     }
 26 }Line[MAXN];
 27 
 28 //X用于离散化横坐标，times为此区间被覆盖的次数，block为有多少块子区间， len为被覆盖的长度
 29 int X[MAXN<<1], times[MAXN<<2], block[MAXN<<2], len[MAXN];
 30 bool usedl[MAXN<<2], usedr[MAXN<<2];
 31 //usedl用于表示区间的左端是否被覆盖， usedr亦如此
 32 
 33 void push_up(int u, int l, int r)
 34 {
 35     if(times[u]>0) //该区间有被覆盖
 36     {
 37         len[u] = X[r] - X[l];
 38         block[u] = 1;
 39         usedl[u] = usedr[u] = true;
 40     }
 41     else    //该区间没有被覆盖
 42     {
 43         if(l+1==r)  //该区间为单位区间
 44         {
 45             len[u] = 0;
 46             block[u] = 0;
 47             usedl[u] = usedr[u] = false;
 48         }
 49         else      //该区间至少包含两个单位区间
 50         {
 51             len[u] = len[u*2] + len[u*2+1];
 52             block[u] = block[u*2] + block[u*2+1];
 53             if(usedr[u*2] && usedl[u*2+1]) //如果左半区间的右端与右半区间的左端均被覆盖，则他们合成一个子区间
 54                 block[u]--;
 55             usedl[u] = usedl[u*2];
 56             usedr[u] = usedr[u*2+1];
 57         }
 58     }
 59 }
 60 
 61 void add(int u, int l, int r, int x, int y, int v)
 62 {
 63     if(x<=l && r<=y)
 64     {
 65         times[u] += v;
 66         push_up(u, l, r);
 67         return;
 68     }
 69 
 70     int mid = (l+r)>>1;
 71     if(x<=mid-1) add(u*2, l, mid, x, y, v);
 72     if(y>=mid+1) add(u*2+1, mid, r, x, y, v);
 73     push_up(u, l, r);
 74 }
 75 
 76 int main()
 77 {
 78     int n;
 79     while(scanf("%d", &n)!=EOF)
 80     {
 81         for(int i = 1; i<=n; i++)
 82         {
 83             int x1, y1, x2, y2;
 84             scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
 85             Line[i].le = Line[i+n].le = x1;
 86             Line[i].ri = Line[i+n].ri = x2;
 87             Line[i].h = y1; Line[i+n].h = y2;
 88             Line[i].id = 1; Line[i+n].id = -1;
 89             X[i] = x1; X[i+n] = x2;
 90         }
 91 
 92         sort(Line+1, Line+1+2*n);
 93         sort(X+1, X+1+2*n);
 94         int m = unique(X+1, X+1+2*n) - (X+1);
 95 
 96         memset(times, 0, sizeof(times));
 97         memset(len, 0, sizeof(len));
 98         memset(block, 0, sizeof(block));
 99         memset(usedl, false, sizeof(usedl));
100         memset(usedr, false, sizeof(usedr));
101 
102         int ans = 0, pre_len = 0;
103         Line[2*n+1].h = Line[2*n].h;    //边界条件
104         for(int i = 1; i<=2*n; i++)
105         {
106             int l = upper_bound(X+1, X+1+m, Line[i].le) - (X+1);
107             int r = upper_bound(X+1, X+1+m, Line[i].ri) - (X+1);
108             add(1, 1, m, l, r, Line[i].id);
109             ans += abs(len[1] - pre_len);
110             ans += 2*block[1]*(Line[i+1].h-Line[i].h);
111             pre_len = len[1];
112         }
113 
114         printf("%d\n", ans);
115     }
116 }