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：

 #include <set>
 #include <map>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <bitset>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <cstdlib>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 typedef long long ll;
 typedef unsigned long long ull;

 #define lson i<<1,l,mid
 #define rson i<<1|1,mid+1,r
 #define bug printf("*********\n");
 #define FIN freopen("D://code//in.txt", "r", stdin);
 #define debug(x) cout<<"["<<x<<"]" <<endl;
 #define IO ios::sync_with_stdio(false),cin.tie(0);

 const double eps = 1e-;
 const int mod = ;
 const int maxn =  + ;
 const double pi = acos(-);
 const int inf = 0x3f3f3f3f;
 const ll INF = 0x3f3f3f3f3f3f3f3f;

 inline int read() {//读入挂
     int ret = , c, f = ;
     for(c = getchar(); !(isdigit(c) || c == '-'); c = getchar());
     if(c == '-') f = -, c = getchar();
     for(; isdigit(c); c = getchar()) ret = ret *  + c - '';
     if(f < ) ret = -ret;
     return ret;
 }

 int n, t;
 double x1, y_1, x2, y_2;
 double y[];

 struct Line {
     double y1, y2, x;
     int flag;
     bool operator < (const Line& a) const {
         return x < a.x;
     }
 }line[];

 struct node {
     int l, r, cover;
     double lf, rf, len;
 }segtree[maxn*];

 void push_up(int i) {
     if(segtree[i].cover > ) {
         segtree[i].len = segtree[i].rf - segtree[i].lf;
     } else if(segtree[i].l +  == segtree[i].r) {
         segtree[i].len = ;
     } else {
         segtree[i].len = segtree[i*].len + segtree[i*+].len;
     }
 }

 void build(int i, int l, int r) {
     segtree[i].l = l, segtree[i].r = r;
     segtree[i].lf = y[l], segtree[i].rf = y[r];
     segtree[i].cover = segtree[i].len = ;
     if(l +  == r) return;
     int mid = (l + r) >> ;
     build(i * , l, mid);
     build(i *  + , mid, r);
 }

 void update(int i, double l, double r, int flag) {
     if(segtree[i].lf == l && segtree[i].rf == r) {
         segtree[i].cover += flag;
         push_up(i);
         return;
     }
     if(l >= segtree[i *  + ].lf) {
         update(i *  + , l, r, flag);
     } else if(r <= segtree[i * ].rf) {
         update(i * , l, r, flag);
     } else {
         update(i * , l, segtree[i*].rf, flag);
         update(i *  + , segtree[i*+].lf, r, flag);
     }
     push_up(i);
 }

 int main() {
     //FIN;
     int icase = ;
     while(~scanf("%d", &n) && n) {
         t = ;
         printf("Test case #%d\n", ++icase);
         for(int i = ; i <= n; i++,t++) {
             scanf("%lf%lf%lf%lf", &x1, &y_1, &x2, &y_2);
             line[t].x = x1;
             line[t].y1 = y_1;
             line[t].y2 = y_2;
             line[t].flag = ;
             y[t] = y_1;
             line[++t].x = x2;
             line[t].y1 = y_1;
             line[t].y2 = y_2;
             line[t].flag = -;
             y[t] = y_2;
         }
         sort(line + , line + t);
         sort(y + , y + t);
         build(, , t - );
         double ans = ;
         update(, line[].y1, line[].y2, line[].flag);
         for(int i = ; i < t; i++) {
             ans += segtree[].len * (line[i].x - line[i-].x);
             update(, line[i].y1, line[i].y2, line[i].flag);
         }
         printf("Total explored area: %.2f\n\n", ans);
     }
     return ;
 }