Atlantis（POJ1151+线段树+扫描线）

题目链接：http://poj.org/problem?id=1151

题目：

题意：求所有矩形的面积，重合部分只算一次。

思路：扫描线入门题，推荐几篇学扫描线的博客：

1.http://www.cnblogs.com/scau20110726/archive/2013/04/12/3016765.html

2.https://blog.csdn.net/qq_38786088/article/details/78633478

3.https://blog.csdn.net/lwt36/article/details/48908031（强烈推荐！）

代码实现如下：

  1 #include <set>
  2 #include <map>
  3 #include <queue>
  4 #include <stack>
  5 #include <cmath>
  6 #include <bitset>
  7 #include <cstdio>
  8 #include <string>
  9 #include <vector>
 10 #include <cstdlib>
 11 #include <cstring>
 12 #include <iostream>
 13 #include <algorithm>
 14 using namespace std;
 15 
 16 typedef long long ll;
 17 typedef unsigned long long ull;
 18 
 19 #define lson i<<1,l,mid
 20 #define rson i<<1|1,mid+1,r
 21 #define bug printf("*********\n");
 22 #define FIN freopen("D://code//in.txt", "r", stdin);
 23 #define debug(x) cout<<"["<<x<<"]" <<endl;
 24 #define IO ios::sync_with_stdio(false),cin.tie(0);
 25 
 26 const double eps = 1e-8;
 27 const int mod = 10007;
 28 const int maxn = 200 + 7;
 29 const double pi = acos(-1);
 30 const int inf = 0x3f3f3f3f;
 31 const ll INF = 0x3f3f3f3f3f3f3f3f;
 32 
 33 inline int read() {//读入挂
 34     int ret = 0, c, f = 1;
 35     for(c = getchar(); !(isdigit(c) || c == '-'); c = getchar());
 36     if(c == '-') f = -1, c = getchar();
 37     for(; isdigit(c); c = getchar()) ret = ret * 10 + c - '0';
 38     if(f < 0) ret = -ret;
 39     return ret;
 40 }
 41 
 42 int n, t;
 43 double x1, y_1, x2, y_2;
 44 double y[1007];
 45 
 46 struct Line {
 47     double y1, y2, x;
 48     int flag;
 49     bool operator < (const Line& a) const {
 50         return x < a.x;
 51     }
 52 }line[305];
 53 
 54 struct node {
 55     int l, r, cover;
 56     double lf, rf, len;
 57 }segtree[maxn*4];
 58 
 59 void push_up(int i) {
 60     if(segtree[i].cover > 0) {
 61         segtree[i].len = segtree[i].rf - segtree[i].lf;
 62     } else if(segtree[i].l + 1 == segtree[i].r) {
 63         segtree[i].len = 0;
 64     } else {
 65         segtree[i].len = segtree[i*2].len + segtree[i*2+1].len;
 66     }
 67 }
 68 
 69 void build(int i, int l, int r) {
 70     segtree[i].l = l, segtree[i].r = r;
 71     segtree[i].lf = y[l], segtree[i].rf = y[r];
 72     segtree[i].cover = segtree[i].len = 0;
 73     if(l + 1 == r) return;
 74     int mid = (l + r) >> 1;
 75     build(i * 2, l, mid);
 76     build(i * 2 + 1, mid, r);
 77 }
 78 
 79 void update(int i, double l, double r, int flag) {
 80     if(segtree[i].lf == l && segtree[i].rf == r) {
 81         segtree[i].cover += flag;
 82         push_up(i);
 83         return;
 84     }
 85     if(l >= segtree[i * 2 + 1].lf) {
 86         update(i * 2 + 1, l, r, flag);
 87     } else if(r <= segtree[i * 2].rf) {
 88         update(i * 2, l, r, flag);
 89     } else {
 90         update(i * 2, l, segtree[i*2].rf, flag);
 91         update(i * 2 + 1, segtree[i*2+1].lf, r, flag);
 92     }
 93     push_up(i);
 94 }
 95 
 96 int main() {
 97     //FIN;
 98     int icase = 0;
 99     while(~scanf("%d", &n) && n) {
100         t = 1;
101         printf("Test case #%d\n", ++icase);
102         for(int i = 1; i <= n; i++,t++) {
103             scanf("%lf%lf%lf%lf", &x1, &y_1, &x2, &y_2);
104             line[t].x = x1;
105             line[t].y1 = y_1;
106             line[t].y2 = y_2;
107             line[t].flag = 1;
108             y[t] = y_1;
109             line[++t].x = x2;
110             line[t].y1 = y_1;
111             line[t].y2 = y_2;
112             line[t].flag = -1;
113             y[t] = y_2;
114         }
115         sort(line + 1, line + t);
116         sort(y + 1, y + t);
117         build(1, 1, t - 1);
118         double ans = 0;
119         update(1, line[1].y1, line[1].y2, line[1].flag);
120         for(int i = 2; i < t; i++) {
121             ans += segtree[1].len * (line[i].x - line[i-1].x);
122             update(1, line[i].y1, line[i].y2, line[i].flag);
123         }
124         printf("Total explored area: %.2f\n\n", ans);
125     }
126     return 0;
127 }

 

posted @ 2018-07-26 00:55  Dillonh  阅读(305)  评论(0)  编辑  收藏  举报