hdu 1542 Atlantis(扫描线)

题目链接：hdu 1542 Atlantis

题目大意：N个矩形，求矩形面积的并。

解题思路：线段树扫描线的裸题。

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>

using namespace std;

const int maxn = 1005;

#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)
int lc[maxn << 2], rc[maxn << 2], v[maxn << 2];
double s[maxn << 2];
vector<double> pos;

inline void pushup(int u) {
    if (v[u])
        s[u] = pos[rc[u] + 1] - pos[lc[u]];
    else if (rc[u] == lc[u])
        s[u] = 0;
    else
        s[u] = s[lson(u)] + s[rson(u)];
}

inline void add(int u, int set) {
    v[u] += set;
    pushup(u);
}

void build (int u, int l, int r) {
    lc[u] = l;
    rc[u] = r;
    v[u] = 0;
    s[u] = 0;

    if (l == r)
        return;

    int mid = (l + r) / 2;
    build(lson(u), l, mid);
    build(rson(u), mid + 1, r);
    pushup(u);
}

void modify (int u, int l, int r, int set) {
    if (l <= lc[u] && rc[u] <= r) {
        add(u, set);
        return;
    }

    int mid = (lc[u] + rc[u]) / 2;
    if (l <= mid)
        modify(lson(u), l, r, set);
    if (r > mid)
        modify(rson(u), l, r, set);
    pushup(u);
}

struct segment {
    double x;
    int l, r, set;
    segment (double x = 0, int l = 0, int r = 0, int set = 0) {
        this->x = x;
        this->l = l;
        this->r = r;
        this->set = set;
    }
};

int N;
vector<segment> vec;
double x1[maxn], x2[maxn], y1[maxn], y2[maxn];

inline bool cmp (const segment& a, const segment& b) {
    return a.x < b.x;
}

inline int find (double k) {
    return lower_bound(pos.begin(), pos.end(), k) - pos.begin();
}

int main () {
    int cas = 1;
    while (scanf("%d", &N) == 1 && N) {
        double ans = 0;
        vec.clear();
        pos.clear();

        for (int i = 0; i < N; i++) {
            scanf("%lf%lf%lf%lf", &x1[i], &y1[i], &x2[i], &y2[i]);
            pos.push_back(y1[i]);
            pos.push_back(y2[i]);
        }
        sort(pos.begin(), pos.end());
        build(1, 0, pos.size() * 2 - 1);

        for (int i = 0; i < N; i++) {
            vec.push_back(segment(x1[i], find(y1[i]), find(y2[i]) - 1, 1));
            vec.push_back(segment(x2[i], find(y1[i]), find(y2[i]) - 1, -1));
        }

        sort(vec.begin(), vec.end(), cmp);
        for (int i = 0; i < vec.size() - 1; i++) {
            modify(1, vec[i].l, vec[i].r, vec[i].set);
            ans += (s[1] * (vec[i+1].x - vec[i].x));
        }

        printf("Test case #%d\n", cas++);
        printf("Total explored area: %.2lf\n\n", ans);
    }
    return 0;
}