UVA 11983 Weird Advertisement

UVA_11983

    这个题目可以转化成求被覆盖K次的矩形面积的并，只要将x2和y2都+1即可，相当于我们把每个单位正方形看成一个点。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define MAXK 11
#define MAXD 60010
int N, K, M, len[4 * MAXD][MAXK], cnt[4 * MAXD], ty[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 cmps(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(len[cur], 0, sizeof(len[cur]));
    len[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, x1, y1, x2, y2;
    scanf("%d%d", &N, &K);
    for(i = 0; i < N; i ++)
    {
        j = i << 1, k = (i << 1) | 1;
        scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
        ++ 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]), 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 = (min + max) >> 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(len[cur], 0, sizeof(len[cur]));
    if(cnt[cur] >= K)
        len[cur][K] = ty[y + 1] - ty[x];
    else if(x == y)
        len[cur][cnt[cur]] = ty[y + 1] - ty[x];
    else
    {
        int i;
        for(i = cnt[cur]; i <= K; i ++)
            len[cur][i] += len[ls][i - cnt[cur]] + len[rs][i - cnt[cur]];
        for(i = K - cnt[cur] + 1; i <= K; i ++)
            len[cur][K] += len[ls][i] + len[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, k;
    long long int 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);
        refresh(1, 0, M - 1, j, k - 1, seg[i].col);
        ans += (long long int)len[1][K] * (seg[i + 1].x - seg[i].x);
    }
    printf("%lld\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;
}