Atlantis HDU

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.

InputThe 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.OutputFor 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

2
10 10 20 20
15 15 25 25.5
0

Sample Output

Test case #1
Total explored area: 180.00 

题解：线段树维护的是算面积时的有效宽。

明白了线段树维护的是啥，剩下的就是理解代码了。。。感谢前辈们的贡献。

#pragma warning(disable:4996)
#include<cstdio>
#include<stack>
#include<cmath>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

#define lson root<<1
#define rson root<<1|1
#define ll long long

const int maxn = 666;

int n, cnt[maxn];
double Hash[maxn], sum[maxn];

struct Seg {
    int d;
    double l, r, h;
    Seg(){}
    Seg(double l, double r, double h, int d): l(l), r(r), h(h), d(d) {}
    bool operator<(const Seg& rhs)const { return h < rhs.h; }
}line[maxn];

void Push_up(int l, int r, int root) {
    if (cnt[root]) sum[root] = Hash[r] - Hash[l];       //如果区间被全部覆盖了，直接赋值
    else if (l + 1 == r) sum[root] = 0;                 //如果区间没有被覆盖且是叶子区间，直接赋0
    else sum[root] = sum[lson] + sum[rson];             //如果区间没有被覆盖但不是叶子区间。为什么这样做呢？因为它的子区间可能被覆盖了，所以等于左区间加上右区间。
}

void Update(int L, int R, int l, int r, int root, int c) {
    if (l >= R || r <= L) return;                       //和点组成的线段树不同（点组成的线段树：if(l>R || r<L) return;）
    if (L <= l && r <= R) {
        cnt[root] += c;
        Push_up(l, r, root);                           
        return;
    }
    int mid = (l + r) >> 1;
    Update(L, R, l, mid, lson, c);
    Update(L, R, mid, r, rson, c);
    Push_up(l, r, root);
}

int main()
{
    int kase = 0;
    while (scanf("%d", &n) && n) {
        double x1, y1, x2, y2;
        for (int i = 1; i <= n; i++) {
            scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
            line[i] = Seg(x1, x2, y1, 1);
            line[i + n] = Seg(x1, x2, y2, -1);
            Hash[i] = x1;
            Hash[i + n] = x2;
        }
        n <<= 1;
        sort(line + 1, line + n + 1);
        sort(Hash + 1, Hash + n + 1);
        
        int m = unique(Hash + 1, Hash + n + 1) - (Hash + 1);


        memset(sum, 0, sizeof(sum));
        memset(cnt, 0, sizeof(cnt));

        double ans = 0;
        for (int i = 1; i < n; i++) {
            int l = lower_bound(Hash + 1, Hash + m + 1, line[i].l) - Hash;
            int r = lower_bound(Hash + 1, Hash + m + 1, line[i].r) - Hash;
        
            Update(l, r, 1, m, 1, line[i].d);
            ans += sum[1] * (line[i + 1].h - line[i].h);
        }
        printf("Test case #%d
Total explored area: %.2lf

", ++kase, ans);
    }
    return 0;
}