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;
}