hdu1542(掃描線求面積並)

題意：求多個矩形面積的並

思路：掃描線算法，對每個矩形，用兩條平行於x軸的線段表示，然後對每條線段的y排序，然後就用平行於x軸的掃描線從y = 0出往上掃，每次求面積。這裏n只有100，所以可以用暴力解。

代碼如下:

#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<stdio.h>
#include<math.h>

#define N 1100
#define inf 0x7ffffff
#define eps 1e-9
#define pi acos(-1.0)
using namespace std;
double x[2*N];
struct node{
    double x1,x2,h;
    int flag;
    bool operator < (const node &a)const
    {
        if(h != a.h)  return h < a.h;
        return flag < a.flag;
    }
}s[N*2];
int color[2*N],b[N*2];
int main()
{
//freopen("input.txt","r",stdin);
//freopen("output.txt","w",stdout);
    int n,cas = 1;
    while(scanf("%d",&n) && n)
    {
        double x1,x2,y1,y2;
        int i;
        for(i = 0; i < n; i++)
        {
            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
            x[2*i] = x1;
            x[2*i+1] = x2;

            s[2*i].x1 = x1;
            s[2*i].x2 = x2;
            s[2*i].h = y1;
            s[2*i].flag = 1;

            s[2*i+1].x1 = x1;
            s[2*i+1].x2 = x2;
            s[2*i+1].h = y2;
            s[2*i+1].flag = -1;
        }
        sort(s,s+2*n);
        sort(x,x+2*n);
        int tot = unique(x,x+2*n) - x;

        memset(color,0,sizeof(color));
        double ans = 0;

        for(i = 0; i < 2*n-1; i++)
        {
            int l = lower_bound(x,x+tot,s[i].x1 ) - x;
            int r = lower_bound(x,x+tot,s[i].x2 ) - x;

            for(int j = 2*l; j <= 2*r; j++)//注意要擴大兩倍，不然有些數據過不了
                color[j] += s[i].flag;
            //for(int j = 0; j < tot; j++)    cout<<color[j]<<" ";cout<<endl;

            int k = 0;//我是用暴力解的，也可以用線段樹維護底邊長
            if(color[0])    b[k++] = 0;
            for(int j = 0; j < 2*tot; j++)
                if(color[j] == 0 && color[j+1]) b[k++] = j+1;
                else if(color[j] && color[j+1] == 0)    b[k++] = j;

            double temp = 0;
            for(int j = 0; j < k; j+=2)
                temp += ( x[ b[j+1]/2 ] - x[ b[j]/2 ]);
            ans += temp* (s[i+1].h - s[i].h);

        }
        printf("Test case #%d\n",cas++);
        printf("Total explored area: %.2lf\n\n",ans);
    }
    return 0;
}