线段树典型例题--poj2482

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本文介绍了一种算法，用于解决寻找天空中特定大小矩形区域内星星总亮度最大化的问题。该算法利用离散化加扫描线的方法，并通过线段树维护星星亮度的累积和最大和。

【题意】

Assume the sky is a flat plane. All the stars lie on it with a location (x, y). for each star, there is a grade ranging from 1 to 100, representing its brightness, where 100 is the brightest and 1 is the weakest. The window is a rectangle whose edges are parallel to the x-axis or y-axis. Your task is to tell where I should put the window in order to maximize the sum of the brightness of the stars within the window. Note, the stars which are right on the edge of the window does not count. The window can be translated but rotation is not allowed.

There are at least 1 and at most 10000 stars in the sky. 1<=W，H<=1000000, 0<=x，y<2^31.

【题解】

首先是离散化+扫描线，但线段树如何维护？

将星星拆成两个，(x[i],y[i],light[i])和(x[i],y[i]+h,-light[i])，然后维护两个量，sum和maxsum。sum是前缀和，maxsum是最大的前缀和

具体维护看代码

【代码】

#include <iostream>
#include <cstring>
#include <string>
#include <cstdio>
#include <algorithm>
using namespace std;

const int N=10005;

struct node
{
    int x,y,idx1,idx2,c;
}a[N];

int ss[N*10],ms[N*10];
long long g[N*3];

void ins(int i,int l,int r,int x,int cc)
{
    if (l==r)
    {
        ss[i]+=cc;
        ms[i]+=cc;
        return;
    }
    int mid=(l+r)/2;
    if (x<=mid) ins(i*2,l,mid,x,cc);
    else ins(i*2+1,mid+1,r,x,cc);
    ss[i]=ss[i*2]+ss[i*2+1];
    ms[i]=max(max(ms[i*2],ss[i*2]+ms[i*2+1]),ss[i*2]);
}

bool cmp(node a,node b)
{
    return a.x<b.x || (a.x==b.x && a.y<b.y);
}

int main()
{
    int tot,xl,xr,i,tmp,n,w,h,ans;

    freopen("in2","r",stdin);
    while (scanf("%d%d%d",&n,&w,&h)!=EOF)
    {
        memset(ss,0,sizeof(ss));
        memset(ms,0,sizeof(ms));
        tot=ans=0;
        for (i=0;i<n;i++)
        {
            scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].c);
            g[tot++]=a[i].y;
            g[tot++]=(long long)a[i].y+h;
        }
        sort(g,g+tot);
        tot=unique(g,g+tot)-g;
        sort(a,a+n,cmp);
        for (i=0;i<n;i++)
        {
            a[i].idx1=lower_bound(g,g+tot,a[i].y)-g+1;
            a[i].idx2=lower_bound(g,g+tot,(long long)a[i].y+h)-g+1;
        }
        xl=xr=0;
        while (xr<n && a[xr].x-a[xl].x<w)
        {
            ins(1,1,tot,a[xr].idx1,a[xr].c);
            ins(1,1,tot,a[xr].idx2,-a[xr].c);
            xr++;
        }
        xr--;
        while (xl<=xr)
        {
            ans=max(ans,ms[1]);
            tmp=xl;
            while (xl<=xr && a[xl].x==a[tmp].x)
            {
                ins(1,1,tot,a[xl].idx1,-a[xl].c);
                ins(1,1,tot,a[xl].idx2,a[xl].c);
                xl++;
            }
            while (xr+1<n && a[xr+1].x-a[xl].x<w)
            {
                xr++;
                ins(1,1,tot,a[xr].idx1,a[xr].c);
                ins(1,1,tot,a[xr].idx2,-a[xr].c);
            }
        }
        printf("%d\n",ans);
    }
}