HDU 3624 Get The Treasury

HDU_3624

    这个题目要求去求覆盖三次及以上的部分的体积。

    如果我们把z坐标离散化的话，每一层z中如果某个部分被覆盖了三次及以上的话，那么等价于这一层z在xy平面上的对应的投影被覆盖了三次及以上，因此对于每一层z我们可以先求出投影中被覆盖了三次及以上的面积，然后乘以这一层z的高度就是这一层z中被覆盖了三次及以上的体积了。

#include<stdio.h>

#include<string.h>

#include<stdlib.h>

#define MAXD 2010

#define K 3

int N, M, Z, S, ty[MAXD], tz[MAXD], cover[4 * MAXD][4], cnt[4 * MAXD];

struct Rec

{

    int x1, y1, z1, x2, y2, z2;

}rec[MAXD];

struct Seg

{

    int x, y1, y2, col;

}seg[MAXD];

int cmpint(const void *_p, const void *_q)

{

    int *p = (int *)_p, *q = (int *)_q;

    return *p < *q ? -1 : 1;

}

int cmpseg(const void *_p, const void *_q)

{

    Seg *p = (Seg *)_p, *q = (Seg *)_q;

    return p->x < q->x ? -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;

    scanf("%d", &N);

    for(i = 0; i < N; i ++)

    {

        scanf("%d%d%d%d%d%d", &rec[i].x1, &rec[i].y1, &rec[i].z1, &rec[i].x2, &rec[i].y2, &rec[i].z2);

        tz[i << 1] = rec[i].z1, tz[(i << 1) | 1] = rec[i].z2;

        ty[i << 1] = rec[i].y1, ty[(i << 1) | 1] = rec[i].y2;

    }

    qsort(tz, N << 1, sizeof(tz[0]), cmpint);

    Z = -1;

    for(i = 0; i < (N << 1); i ++)

        if(i == 0 || tz[i] != tz[i - 1])

            tz[++ Z] = tz[i];

    qsort(ty, N << 1, sizeof(ty[0]), cmpint);

    M = -1;

    for(i = 0; i < (N << 1); i ++)

        if(i == 0 || ty[i] != ty[i - 1])

            ty[++ M] = ty[i];

    build(1, 0, M - 1);

}

int BS(int x)

{

    int mid, min = 0, max = M + 1;

    for(;;)

    {

        mid = (max + min) >> 1;

        if(mid == min)

            break;

        if(ty[mid] <= x)

            min = mid;

        else

            max = mid;

    }

    return mid;

}

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);

}

void solve()

{

    int i, j, l, r;

    long long int ans = 0, temp;

    for(i = 0; i < Z; i ++)

    {

        S = 0;

        for(j = 0; j < N; j ++)

            if(rec[j].z1 <= tz[i] && rec[j].z2 >= tz[i + 1])

            {

                seg[S].x = rec[j].x1, seg[S].y1 = rec[j].y1, seg[S].y2 = rec[j].y2, seg[S].col = 1;

                ++ S;

                seg[S].x = rec[j].x2, seg[S].y1 = rec[j].y1, seg[S].y2 = rec[j].y2, seg[S].col = -1;

                ++ S;

            }

        qsort(seg, S, sizeof(seg[0]), cmpseg);

        seg[S].x = seg[S - 1].x;

        temp = 0;

        for(j = 0; j < S; j ++)

        {

            l = BS(seg[j].y1), r = BS(seg[j].y2);

            refresh(1, 0, M - 1, l, r - 1, seg[j].col);

            temp += (long long int)cover[1][3] * (seg[j + 1].x - seg[j].x);

        }

        ans += temp * (tz[i + 1] - tz[i]);

    }

    printf("%I64d\n", ans);

}

int main()

{

    int t, tt;

    scanf("%d", &t);

    for(tt = 0; tt < t; tt ++)

    {

        init();

        printf("Case %d: ", tt + 1);

        solve();

    }

    return 0;

}