HDU-1542 Atlantis 求矩形面积并

http://acm.hdu.edu.cn/showproblem.php?pid=1542

该题就是给定N个矩形，求出叠加之后的面积，与求面积交类似，就是覆盖次数没有了要求。

代码如下：

#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <map>
#include <algorithm>
#define MAXN 205
using namespace std;

struct Node
{
    double x, y1, y2;
    int lr;
    bool operator < (Node t) const
    {
        return t.x-x>1e-6;
    }
}e[MAXN*2];

struct
{
    int l, r;
    int cover;
}seg[MAXN*3];

double yy[MAXN*2];

void creat(int f, int l, int r)
{
    int mid = (l+r)>>1;
    seg[f].l = l, seg[f].r = r;
    seg[f].cover = 0;
    if (r - l > 1)
    {
        creat(f<<1, l, mid);
        creat(f<<1|1, mid, r);
    }
}

void modify(int f, int l, int r, int val)
{
    int mid = (seg[f].l + seg[f].r) >> 1;
    if (seg[f].l == l && seg[f].r == r)
    {
        seg[f].cover += val;
    }
    else if (seg[f].r - seg[f].l > 1)
    {
        if (r <= mid)
            modify(f<<1, l, r, val);
        else if (l >= mid)
            modify(f<<1|1, l, r, val);
        else
        {
            modify(f<<1, l, mid, val);
            modify(f<<1|1, mid, r, val);
        }
    }
}

void query(int f, double &ans)
{
    if (seg[f].cover > 0)
    {
        ans += yy[seg[f].r] - yy[seg[f].l];
    }
    else if (seg[f].r - seg[f].l > 1)
    {
        query(f<<1, ans);
        query(f<<1|1, ans);
    }
}

int main()
{
    double x1, y1, x2, y2, ans, res;
    int N, ca = 1;;
    while (scanf("%d", &N), N)
    {
        res = 0;
        map<double,int>mp;
        for (int i = 1, j = 1; i <= N; ++i, j += 2)
        {
            scanf("%lf %lf %lf %lf", &x1, &y1, &x2, &y2);
            e[j].x = x1, e[j].y1 = y1, e[j].y2 = y2; 
            e[j].lr = 1;
            e[j+1].x = x2, e[j+1].y1 = y1, e[j+1].y2 = y2;
            e[j+1].lr = -1;
            yy[j] = y1, yy[j+1] = y2;
        }
        sort(e+1, e+1+2*N);
        sort(yy+1, yy+1+2*N);
        int cnt = unique(yy+1, yy+1+2*N) - (yy+1);
        creat(1, 1, cnt);
        for (int i = 1; i <= cnt; ++i)
        {
            mp[yy[i]] = i;
        //    printf("yy= %lf\n", yy[i]);
        }
        for (int i = 1; i < 2*N; ++i)
        {
            ans = 0;
            modify(1, mp[e[i].y1], mp[e[i].y2], e[i].lr);
            query(1, ans);
            res += ans * (e[i+1].x - e[i].x);
        //    printf("res = %lf\n", res);
        }
        printf("Test case #%d\nTotal explored area: %.2lf\n\n", ca++, res);
    } 
    return 0;
}