hdu1542 Atlantis（扫描线+矩形面积并）

Atlantis

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14664    Accepted Submission(s): 6045

Problem Description

There are several ancient Greek texts that contain descriptions of the fabled island Atlantis. Some of these texts even include maps of parts of the island. But unfortunately, these maps describe different regions of Atlantis. Your friend Bill has to know the total area for which maps exist. You (unwisely) volunteered to write a program that calculates this quantity.

 

Input

The input file consists of several test cases. Each test case starts with a line containing a single integer n (1<=n<=100) of available maps. The n following lines describe one map each. Each of these lines contains four numbers x1;y1;x2;y2 (0<=x1<x2<=100000;0<=y1<y2<=100000), not necessarily integers. The values (x1; y1) and (x2;y2) are the coordinates of the top-left resp. bottom-right corner of the mapped area.

The input file is terminated by a line containing a single 0. Don’t process it.

 

Output

For each test case, your program should output one section. The first line of each section must be “Test case #k”, where k is the number of the test case (starting with 1). The second one must be “Total explored area: a”, where a is the total explored area (i.e. the area of the union of all rectangles in this test case), printed exact to two digits to the right of the decimal point.

Output a blank line after each test case.

 

Sample Input

210 10 20 2015 15 25 25.50

 

Sample Output

Test case #1Total explored area: 180.00 

 

Source

Mid-Central European Regional Contest 2000

 

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思路:浮点数先要离散化;然后把矩形分成两条边,上边和下边,对横轴建树,然后从下到上扫描上去,用cnt表示该区间下边比上边多几个

线段树操作:update:区间增减 query:直接取根节点的值

#include <cstdio>#include <cstring>#include <cctype>#include <algorithm>using namespace std;#define lson l , m , rt << 1#define rson m + 1 , r , rt << 1 | 1const int maxn = 2222;int cnt[maxn << 2];//边重复的次数double sum[maxn << 2];double X[maxn];struct Seg {double h, l, r;int s;//1为入边，-1为出边Seg() {}Seg(double a, double b, double c, int d) : l(a), r(b), h(c), s(d) {}bool operator < (const Seg &cmp) const {return h < cmp.h;}}ss[maxn];void PushUp(int rt, int l, int r) {//每个区间求[l, r-1]，算的时候右边加一，这样递归的时候就不用考虑叶子节点了。if (cnt[rt]) sum[rt] = X[r + 1] - X[l];//这条边存在else if (l == r) sum[rt] = 0;//边不存在，又是叶子节点else sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];}void update(int L, int R, int c, int l, int r, int rt) {if (L <= l && r <= R) {cnt[rt] += c;PushUp(rt, l, r);return;}int m = (l + r) >> 1;if (L <= m) update(L, R, c, lson);if (m < R) update(L, R, c, rson);PushUp(rt, l, r);}int main() {int n, cas = 1;while (~scanf("%d", &n) && n) {int m = 0;while (n--) {double a, b, c, d;scanf("%lf%lf%lf%lf", &a, &b, &c, &d);X[m] = a;ss[m++] = Seg(a, c, b, 1);X[m] = c;ss[m++] = Seg(a, c, d, -1);}//离散化sort(X, X + m);sort(ss, ss + m);int k = unique(X, X + m) - X;memset(cnt, 0, sizeof(cnt));memset(sum, 0, sizeof(sum));double ret = 0;for (int i = 0; i < m - 1; i++) {int l = lower_bound(X, X + k, ss[i].l) - X;int r = lower_bound(X, X + k, ss[i].r) - X - 1;if (l <= r) update(l, r, ss[i].s, 0, k - 1, 1);ret += sum[1] * (ss[i + 1].h - ss[i].h);//sum[1]代表x轴的长度和}printf("Test case #%d\nTotal explored area: %.2lf\n\n", cas++, ret);}return 0;}