HDU 1542 Atlanta (matrix coverage area of Line Segment tree)

Question: Calculate the coverage area for N matrices.

Thought: I read others' answer reports

Given the coordinates in the lower left corner of a rectangle and the coordinates in the upper right corner are respectively: (x1, Y1) and (X2, Y2). For such a rectangle, we construct two line segments, one positioned in X1, in the Y coordinate, the range is [Y1, y2], and the value of a cover domain is 1. The other line segment is located in X2, And the range is [Y1, y2], given a cover value of-1. Based on this method, two line segments are constructed for each rectangle, and all line segments are sorted from left to right based on the located X.

# Include <iostream> # include <stdio. h> # include <string. h >#include <algorithm> using namespace STD; # define M 100 # define INF 0x3fffffff # define maxn 500000*2 struct seg {int flag; double up, down, X ;} line [M * 5]; int CMP (seg a, seg B) {return. x <B. x;} struct tree {double up, down, X; int cover; bool flag;} tree [M * m]; Double Y [M * 3]; // split y and build void build (int id, int L, int R) {tree [ID]. down = Y [l]; tree [ID]. up = Y [R]; tree [ID]. flag = false; tree [ID]. cover = 0; tree [ID]. X =-1; if (L + 1 = r) {tree [ID]. flag = true; return;} int mid = (L + r)/2; build (ID * 2, L, mid); Build (ID * 2 + 1, mid, r);} double insert (int id, double X, double L, Double R, int flag) // flag indicates left or right {If (tree [ID]. down> = r | tree [ID]. up <= L) return 0; If (tree [ID]. flag) {If (tree [ID]. cover> 0) // Recursion to the leaf node {double temp_x = tree [ID]. x; double ans = (x-temp_x) * (tree [ID]. up-tree [ID]. down); tree [ID]. X = x; // locate the last xtree [ID]. cover + = flag; return ans;} else {tree [ID]. cover + = flag; tree [ID]. X = x; return 0 ;}} double ans1, ans2; ans1 = insert (ID * 2, X, L, R, flag ); ans2 = insert (ID * 2 + 1, x, L, R, flag); Return ans1 + ans2;} int main () {int T, CA = 1; while (scanf ("% d", & T) {double x1, x2, Y1, Y2; int K = 0; For (INT I = 0; I <t; I ++) {scanf ("% lf", & X1, & Y1, & X2, & Y2); k ++; Y [k] = Y1; line [K]. down = Y1; line [K]. up = Y2; line [K]. X = x1; line [K]. flag = 1; // 1 indicates K ++ on the left; y [k] = Y2; line [K]. X = x2; line [K]. down = Y1; line [K]. up = Y2; line [K]. flag =-1; //-1 indicates the right} Sort (Y + 1, Y + k + 1); sort (LINE + 1, line + 1 + k, CMP ); build (1, 1, k); double ans = 0; For (INT I = 1; I <= K; I ++) ans + = insert (1, line [I]. x, line [I]. down, line [I]. up, line [I]. flag); printf ("Test Case # % d \ ntotal received area: %. 2f \ n ", CA ++, ANS);} return 0;}/* 210 10 20 2015 15 25 25.50 */