跟pku1151是一樣的。
題目給出 n 個(gè)矩形，要求它們的面積并。具體做法是離散化。先把 2n 個(gè) x 坐標(biāo)排序去重，然后再把所有水平線段（要記錄是矩形上邊還是下邊）按 y 坐標(biāo)排序。最后對于每一小段區(qū)間 (x[i], x[i + 1]) 掃描所有的水平線段，求出這些水平線段在小區(qū)間內(nèi)覆蓋的面積。總的時(shí)間復(fù)雜度是 O(n
2)。利用線段樹，可以優(yōu)化到 O(nlogn)。
/**//*************************************************************************
Author: WHU_GCC
Created Time: 2007-8-25 17:41:41
File Name: pku1389.cpp
Description:
************************************************************************/
#include <iostream>
#include <cmath>
using namespace std;
#define out(x) (cout << #x << ": " << x << endl)
const int maxint = 0x7FFFFFFF;
typedef long long int64;
const int64 maxint64 = 0x7FFFFFFFFFFFFFFFLL;
template <class T> void show(T a, int n) 
{for (int i = 0; i < n; ++i) cout << a[i] << ' '; cout << endl; }
template <class T> void show(T a, int r, int l) {for (int i = 0; i < r; ++i) show(a[i], l); cout << endl; }
const int maxn = 4010;
typedef struct line_t
{
double l, r, y;
int flag;
};
bool operator <(const line_t &a, const line_t &b)
{
return a.y < b.y;
}
bool d_equal(const double &a, const double &b)
{
return abs(a - b) < 1e-9;
}
int n;
double x[maxn];
int cnt_x, cnt_line;
line_t line[maxn];
int main()
{
while (1)
{
cnt_x = 0;
cnt_line = 0;
while (1)
{
double x1, y1, x2, y2;
scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
if (x1 == -1.0) break;
x[cnt_x++] = x1;
x[cnt_x++] = x2;
line[cnt_line].flag = 0;
line[cnt_line].l = x1;
line[cnt_line].r = x2;
line[cnt_line++].y = y1;
line[cnt_line].flag = 1;
line[cnt_line].l = x1;
line[cnt_line].r = x2;
line[cnt_line++].y = y2;
}
sort(line, line + cnt_line);
sort(x, x + cnt_x);
cnt_x = unique(x, x + cnt_x, d_equal) - x;
if (cnt_x == 0) break;
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[i] && line[j].r >= x[i + 1])
{
if (cnt == 0) now_y = line[j].y;
if (line[j].flag == 0) cnt++;
else cnt--;
if (cnt == 0) area += (x[i + 1] - x[i]) * (line[j].y - now_y);
}
}
printf("%.0lf\n", area);
}
return 0;
}