学步园

HDU_1225

    拓展一下求K次覆盖的矩形的并的话就是UVA_11983那个题了。感觉上用线段树处理矩形的并，首先就是要标记出哪些区间被覆盖了，其次就是要用类似dp的思想，用cover[i][j]表示在线段树上的节点i表示的范围内，覆盖了j次的线段总长度，同时cover[i][K]表示的是覆盖了K次及大于K次的线段的总长度，然后依据左右儿子的状态来更新cover[i][j]的状态。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#define MAXD 2010
#define zero 1e-8
int N, M, cnt[4 * MAXD], K = 2;
double ty[MAXD], cover[4 * MAXD][3];
struct Seg
{
    double x, y1, y2;
    int col;
}seg[MAXD];
int cmpy(const void *_p, const void *_q)
{
    double *p = (double *)_p, *q = (double *)_q;
    return *p < *q ? -1 : 1;
}
int cmps(const void *_p, const void *_q)
{
    Seg *p = (Seg *)_p, *q = (Seg *)_q;
    return p->x < q->x ? -1 : 1;
}
int dcmp(double x)
{
    return fabs(x) < zero ? 0 : (x < 0 ? -1 : 1);
}
void build(int cur, int x, int y)
{
    int mid = (x + y) >> 1, ls = cur << 1, rs = (cur << 1) | 1;
    memset(cover[cur], 0, sizeof(cover[cur]));
    cover[cur][0] = ty[y + 1] - ty[x];
    cnt[cur] = 0;
    if(x == y)
        return ;
    build(ls, x, mid);
    build(rs, mid + 1, y);
}
void init()
{
    int i, j, k;
    double x1, x2, y1, y2;
    scanf("%d", &N);
    for(i = 0; i < N; i ++)
    {
        j = i << 1, k = (i << 1) | 1;
        scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
        seg[j].x = x1, seg[k].x = x2;
        seg[j].y1 = seg[k].y1 = y1, seg[j].y2 = seg[k].y2 = y2;
        seg[j].col = 1, seg[k].col = -1;
        ty[j] = y1, ty[k] = y2;
    }
    qsort(ty, N << 1, sizeof(ty[0]), cmpy);
    M = (N << 1) - 1;
    build(1, 0, M - 1);
}
void update(int cur, int x, int y)
{
    int ls = cur << 1, rs = (cur << 1) | 1;
    memset(cover[cur], 0, sizeof(cover[cur]));
    if(cnt[cur] >= K)
        cover[cur][K] = ty[y + 1] - ty[x];
    else if(x == y)
        cover[cur][cnt[cur]] = ty[y + 1] - ty[x];
    else
    {
        int i;
        for(i = cnt[cur]; i <= K; i ++)
            cover[cur][i] += cover[ls][i - cnt[cur]] + cover[rs][i - cnt[cur]];
        for(i = K - cnt[cur] + 1; i <= K; i ++)
            cover[cur][K] += cover[ls][i] + cover[rs][i];
    }
}
void refresh(int cur, int x, int y, int s, int t, int c)
{
    int mid = (x + y) >> 1, ls = cur << 1, rs = (cur << 1) | 1;
    if(x >= s && y <= t)
    {
        cnt[cur] += c;
        update(cur, x, y);
        return ;
    }
    if(mid >= s)
        refresh(ls, x, mid, s, t, c);
    if(mid + 1 <= t)
        refresh(rs, mid + 1, y, s, t, c);
    update(cur, x, y);
}
int BS(double x)
{
    int min = 0, max = M + 1, mid;
    for(;;)
    {
        mid = (min + max) >> 1;
        if(mid == min)
            break;
        if(dcmp(ty[mid] - x) <= 0)
            min = mid;
        else
            max = mid;
    }
    return mid;
}
void solve()
{
    int i, j, k;
    double ans = 0;
    qsort(seg, N << 1, sizeof(seg[0]), cmps);
    seg[N << 1].x = seg[(N << 1) - 1].x;
    for(i = 0; i < (N << 1); i ++)
    {
        j = BS(seg[i].y1), k = BS(seg[i].y2);
        if(j < k)
            refresh(1, 0, M - 1, j, k - 1, seg[i].col);
        ans += cover[1][K] * (seg[i + 1].x - seg[i].x);
    }
    printf("%.2f\n", ans);
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t --)
    {
        init();
        solve();
    }
    return 0;
}