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("*********
");
#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-8;
const int mod = 10007;
const int maxn = 200 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;

inline int read() {//读入挂
    int ret = 0, c, f = 1;
    for(c = getchar(); !(isdigit(c) || c == '-'); c = getchar());
    if(c == '-') f = -1, c = getchar();
    for(; isdigit(c); c = getchar()) ret = ret * 10 + c - '0';
    if(f < 0) ret = -ret;
    return ret;
}

int n, t;
double x1, y_1, x2, y_2;
double y[1007];

struct Line {
    double y1, y2, x;
    int flag;
    bool operator < (const Line& a) const {
        return x < a.x;
    }
}line[305];

struct node {
    int l, r, cover;
    double lf, rf, len;
}segtree[maxn*4];

void push_up(int i) {
    if(segtree[i].cover > 0) {
        segtree[i].len = segtree[i].rf - segtree[i].lf;
    } else if(segtree[i].l + 1 == segtree[i].r) {
        segtree[i].len = 0;
    } else {
        segtree[i].len = segtree[i*2].len + segtree[i*2+1].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 = 0;
    if(l + 1 == r) return;
    int mid = (l + r) >> 1;
    build(i * 2, l, mid);
    build(i * 2 + 1, 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 * 2 + 1].lf) {
        update(i * 2 + 1, l, r, flag);
    } else if(r <= segtree[i * 2].rf) {
        update(i * 2, l, r, flag);
    } else {
        update(i * 2, l, segtree[i*2].rf, flag);
        update(i * 2 + 1, segtree[i*2+1].lf, r, flag);
    }
    push_up(i);
}

int main() {
    //FIN;
    int icase = 0;
    while(~scanf("%d", &n) && n) {
        t = 1;
        printf("Test case #%d
", ++icase);
        for(int i = 1; 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 = 1;
            y[t] = y_1;
            line[++t].x = x2;
            line[t].y1 = y_1;
            line[t].y2 = y_2;
            line[t].flag = -1;
            y[t] = y_2;
        }
        sort(line + 1, line + t);
        sort(y + 1, y + t);
        build(1, 1, t - 1);
        double ans = 0;
        update(1, line[1].y1, line[1].y2, line[1].flag);
        for(int i = 2; i < t; i++) {
            ans += segtree[1].len * (line[i].x - line[i-1].x);
            update(1, line[i].y1, line[i].y2, line[i].flag);
        }
        printf("Total explored area: %.2f

", ans);
    }
    return 0;
}