poj1151 Atlantis 【离散化+扫描线】

wuxu — Sat, 08 Oct 2011 02:59:00 GMT

求 n (1 <= n <= 100) 个矩形的面积并。题目链接：http://poj.org/problem?id=1151

#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;

const int maxn = 210;
const double eps = 1e-9;

struct line_t
{
     double l_x, r_x, y;
     bool flag;
};

bool cmp(const line_t &a, const line_t &b)
{
     return a.y < b.y;
}

bool is_equal(const double &a, const double &b)
{
     return abs(a - b) < eps;
}

int n;
double x[maxn];
int cnt_x, cnt_line;
line_t line[maxn];

int main()
{
     int num = 0;
     while(scanf("%d", &n) != EOF && n)
     {
         cnt_x = 0;
         cnt_line = 0;
         for(int i = 0; i < n; ++i)
         {
             double x1, y1, x2, y2;
             scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
             x[cnt_x++] = x1;
             x[cnt_x++] = x2;
             line[cnt_line].flag = false;
             line[cnt_line].l_x = x1;
             line[cnt_line].r_x = x2;
             line[cnt_line++].y = y1;
             line[cnt_line].flag = true;
             line[cnt_line].l_x = x1;
             line[cnt_line].r_x = x2;
             line[cnt_line++].y = y2;
         }
         sort(line,line + cnt_line, cmp);
         sort(x, x + cnt_x);
         cnt_x = unique(x, x + cnt_x, is_equal) - x;

         double area = 0.0;
         for(int i = 0; i < cnt_x - 1; ++i)
         {
             int cnt = 0;
             double now_y;
             for(int j = 0; j < cnt_line; ++j)
             if(line[j].l_x <= x[i] && line[j].r_x >= x[i + 1])
             {
                  if(cnt == 0) now_y = line[j].y;
                  if(!line[j].flag) ++cnt;
                      else --cnt;
                  if(cnt == 0) area += (x[i + 1] - x[i]) * (line[j].y - now_y);
             }
         }
         printf("Test case #%d\n", ++num);
         printf("Total explored area: %.2lf\n\n", area);
     }
     return 0;
}

wuxu 2011-10-08 10:59 发表评论

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poj1151 Atlantis 【离散化+扫描线】