hdu 1542 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): 20693    Accepted Submission(s): 8258

 

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
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

2 10 10 20 20 15 15 25 25.5 0

 

 

Sample Output

Test case #1 Total explored area: 180.00

题目大意：求矩形的面积并

扫描线，线段树每个节点维护该区间内有多长被大于等于1个边覆盖了，这个题目对于lazy标记不需要pushdown，因为pushdown之后很难维护，pushdown原本是用于修改操作的顺序会影响结果，如果和顺序无关的话是不需要pushdown的，每个节点的值为执行以当前节点为根的子树内部的所有lazy标记之后的值。

#include 
#include 
#include 
#include 
using namespace std;
const int N = 1e3 + 10;
#define lson o * 2, l, m
#define rson o * 2 + 1, m + 1, r
#define ls o * 2
#define rs o * 2 + 1

double sum[N * 4];
int add[N * 4];
int n;
vector  pos; 
struct line {
    double a, b, c;
    int op;
    bool operator<(const line& o) const{
        return c < o.c;
    }
};

vector  L; 

void build(int o, int l, int r) {
    sum[o] = add[o] = 0;
    if (l == r) return;
    int m = (l + r) / 2;
    build(lson);
    build(rson);
}

void pushup(int o, int l, int r) {
    if (add[o]) sum[o] = pos[r + 1] - pos[l];
    else if (l == r) sum[o] = 0;
    else sum[o] = sum[ls] + sum[rs];
}

void update(int o, int l, int r, int L, int R, int op) {
    if (L <= l && r <= R) {
        add[o] += op;
        pushup(o, l, r);
        return;
    }
    int m = (l + r) / 2;
    if (L <= m) update(lson, L, R, op);
    if (R > m) update(rson, L, R, op);
    pushup(o, l, r);
}

int main() {
    int cas = 0;
    while (scanf("%d", &n) != EOF) {
        if (n == 0) break;
        cas++;
        pos.clear();
        L.clear();
        double a, b, c, d;
        for (int i = 1; i <= n; i++) {
            scanf("%lf%lf%lf%lf", &a, &b, &c, &d);
            L.push_back({a, c, b, 1});
            L.push_back({a, c, d, -1});
            pos.push_back(a);
            pos.push_back(c);
        }
        sort(L.begin(), L.end());
        sort(pos.begin(), pos.end());
        pos.erase(unique(pos.begin(), pos.end()), pos.end());
        int mx = pos.size() + 5;
        build(1, 0, mx);
        double ans = 0;
        double last = 0;
        for (int i = 0; i < L.size(); i++) {
            ans += sum[1] * (L[i].c - last);
            int x = lower_bound(pos.begin(), pos.end(), L[i].a) - pos.begin();
            int y = lower_bound(pos.begin(), pos.end(), L[i].b) - pos.begin() - 1;
            update(1, 0, mx, x, y, L[i].op);
            last = L[i].c;
        }
        printf("Test case #%d\n", cas);
        printf("Total explored area: %.2lf\n\n", ans);
    }
    return 0;
}