HDU 3265 Posters

HDU_3265

    如果我们把一个掏了洞的矩形切成4个完整的矩形的话，那么这就跟普通的求矩形面积的并的问题是一样的了。

    此外要注意一下中间结果可能会超过int的范围。

#include<stdio.h>

#include<string.h>

#include<stdlib.h>

#define MAXD 50010

int N, D, len[4 * MAXD], cnt[4 * MAXD], E;

struct Seg

{

    int x, y1, y2, col;

}seg[8 * MAXD];

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

{

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

    return p->x < q->x ? -1 : 1;

}

void addseg(int x, int y1, int y2, int c)

{

    seg[E].x = x, seg[E].y1 = y1, seg[E].y2 = y2, seg[E].col = c;

    ++ E;

}

void build(int cur, int x, int y)

{

    int mid = (x + y) >> 1, ls = cur << 1, rs = (cur << 1) | 1;

    len[cur] = cnt[cur] = 0;

    if(x == y)

        return ;

    build(ls, x, mid);

    build(rs, mid + 1, y);

}

void init()

{

    int i, j, k, x1, y1, x2, y2, x3, y3, x4, y4;

    E = 0;

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

    {

        scanf("%d%d%d%d%d%d%d%d", &x1, &y1, &x2, &y2, &x3, &y3, &x4, &y4);

        addseg(x1, y1, y2, 1), addseg(x3, y1, y2, -1);

        addseg(x4, y1, y2, 1), addseg(x2, y1, y2, -1);

        addseg(x3, y4, y2, 1), addseg(x4, y4, y2, -1);

        addseg(x3, y1, y3, 1), addseg(x4, y1, y3, -1);

    }

    D = 50000;

    build(1, 0, D - 1);

}

void update(int cur, int x, int y)

{

    int ls = cur << 1, rs = (cur << 1) | 1;

    if(cnt[cur])

        len[cur] = y - x + 1;

    else if(x == y)

        len[cur] = 0;

    else

        len[cur] = len[ls] + len[rs];

}

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, k;

    long long int ans = 0;

    qsort(seg, E, sizeof(seg[0]), cmps);

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

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

    {

        if(seg[i].y2 > seg[i].y1)

            refresh(1, 0, D - 1, seg[i].y1, seg[i].y2 - 1, seg[i].col);

        ans += (long long int)len[1] * (seg[i + 1].x - seg[i].x);

    }

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

}

int main()

{

    for(;;)

    {

        scanf("%d", &N);

        if(!N)

            break;

        init();

        solve();

    }

    return 0;

}