POJ 2482 Stars in Your Window

POJ_2482

    这个题目类似黑书上的一个例题,在看了黑书的那个例题之后,以及做过前面关于扫面线的相关练习之后,这个题目的思路就不难理解了。

    首先我们要控制矩形横向的位置,完成这一点可以用一个队列,当队尾的x和队首的x之差大于或等于W就让队首的点出队即可。接下来,就是考虑如果矩形从下往上滑动时,在哪个位置所圈的星星的总亮度最高呢?黑书上介绍了算法,我们可以把每个星星按纵坐标的拆成两个点,一个是y,一个是y+H,如果y点亮度为z的话,那么y+H点的亮度就是-z,这样求y方向上的最大前缀和就得到了最高的总亮度,所以我们需要用一棵线段树维护各个前缀和,并能够查询所有前缀和的最大值。

    更多和线段树、扫描线相关的题目可以参考胡浩的博客:http://www.notonlysuccess.com/index.php/segment-tree-complete/

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define MAXD 20010
struct Point
{
    int x, y, z;
}point[MAXD];
int N, M, W, H, max[4 * MAXD], add[4 * MAXD];
long long int ty[MAXD];
int cmpy(const void *_p, const void *_q)
{
    long long int *p = (long long int *)_p, *q = (long long int *)_q;
    return *p < *q ? -1 : 1;
}
int cmpp(const void *_p, const void *_q)
{
    Point *p = (Point *)_p, *q = (Point *)_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;
    max[cur] = add[cur] = 0;
    if(x == y)
        return ;
    build(ls, x, mid);
    build(rs, mid + 1, y);
}
void init()
{
    int i, j, k;
    for(i = 0; i < N; i ++)
    {
        j = i << 1, k = (i << 1) | 1;
        scanf("%d%d%d", &point[i].x, &point[i].y, &point[i].z);
        ty[j] = point[i].y, ty[j + 1] = (long long int)point[i].y + H;
    }
    qsort(ty, N << 1, sizeof(ty[0]), cmpy);
    M = 0;
    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(long long int x)
{
    int min = 0, max = M, mid;
    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 ls = cur << 1, rs = (cur << 1) | 1;
    max[cur] = max[ls] > max[rs] ? max[ls] : max[rs];
}
void pushdown(int cur)
{
    int ls = cur << 1, rs = (cur << 1) | 1;
    if(add[cur])
    {
        add[ls] += add[cur], add[rs] += add[cur];
        max[ls] += add[cur], max[rs] += add[cur];
        add[cur] = 0;
    }
}
void refresh(int cur, int x, int y, int s, int t, int delta)
{
    int mid = (x + y) >> 1, ls = cur << 1, rs = (cur << 1) | 1;
    if(x >= s && y <= t)
    {
        max[cur] += delta, add[cur] += delta;
        return ;
    }
    pushdown(cur);
    if(mid >= s)
        refresh(ls, x, mid, s, t, delta);
    if(mid + 1 <= t)
        refresh(rs, mid + 1, y, s, t, delta);
    update(cur);
}
void solve()
{
    int i, j, k, l, r, ans = 0;
    qsort(point, N, sizeof(point[0]), cmpp);
    for(i = j = 0; i < N; i ++)
    {
        l = BS(point[i].y), r = BS((long long int)point[i].y + H);
        refresh(1, 0, M - 1, l, M - 1, point[i].z), refresh(1, 0, M - 1, r, M - 1, -point[i].z);
        while(point[j].x <= point[i].x - W)
        {
            l = BS(point[j].y), r = BS((long long int)point[j].y + H);
            refresh(1, 0, M - 1, l, M - 1, -point[j].z), refresh(1, 0, M - 1, r, M - 1, point[j].z);
            ++ j;
        }
        if(max[1] > ans)
            ans = max[1];
    }
    printf("%d\n", ans);
}
int main()
{
    while(scanf("%d%d%d", &N, &W, &H) == 3)
    {
        init();
        solve();
    }
    return 0;
}
原文地址:https://www.cnblogs.com/staginner/p/2440572.html