hdu 3255 体积并

http://www.cnblogs.com/kane0526/archive/2013/03/07/2948446.html

http://blog.csdn.net/acdreamers/article/details/11854781

每块蔬菜地种植蔬菜收获的利润为 val=x*y*price。  面积乘以价格，题目的重点转换在于如何确定重叠区域怎么让它种植最贵的蔬菜。

观察利润计算公式 :   x*y*price <==> x*y*h    可以转换为求体积并。

体积并其实和面积并基本一样，将体积的范围记录下来，每一层遍历求面积并，结果就是底面积乘每一层的高之和

#include <iostream>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <stack>
#include <queue>
#include <cctype>
#include <vector>
#include <iterator>
#include <set>
#include <map>
#include <sstream>
using namespace std;
 
#define mem(a,b) memset(a,b,sizeof(a))
#define pf printf
#define sf scanf
#define spf sprintf
#define pb push_back
#define debug printf("!\n")
#define MAXN 160000+5
#define MAX(a,b) a>b?a:b
#define blank pf("\n")
#define LL long long
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
#define pqueue priority_queue
#define INF 0x3f3f3f3f
 
#define ls (rt<<1)
#define rs (rt<<1|1)
 
int n,m;
 
double hh[MAXN],col[MAXN<<],len[MAXN<<];
 
int V[MAXN];
 
struct node
{
    double l,r,x,c;
    int v;
    node(){}
    node(double a,double b,double c,double d,int e):l(a),r(b),x(c),c(d),v(e){}
    bool operator < (const node &b) const
    {
        return x<b.x;
    }
}a[MAXN<<],tmp[MAXN<<];
 
void PushUp(int rt,int l,int r)
{
    if(col[rt])
    {
        len[rt] = hh[r+] - hh[l];
    }
    else if(l==r) len[rt] = ;
    else
    {
        len[rt] = len[ls]+len[rs];
    }
}
 
void update(int val,int L,int R,int l,int r,int rt)
{
    if(L<=l && r<=R)
    {
        col[rt] += val;
        PushUp(rt,l,r);
        return;
    }
    int mid = (l+r)>>;
    if(L <= mid) update(val,L,R,l,mid,ls);
    if(R > mid) update(val,L,R,mid+,r,rs);
    PushUp(rt,l,r);
}
 
int main()
{
    int n,i,j,t,kase=;
    double ans;
    sf("%d",&t);
    while(t--)
    {
        sf("%d%d",&n,&m);
        int v=;
        for(i=;i<=m;i++)
        sf("%d",&V[i]);
        for(i=;i<=n;i++)
        {
            double x1,x2,y1,y2;
            int r;
            sf("%lf%lf%lf%lf%d",&x1,&y1,&x2,&y2,&r);
            hh[++v]=y1;
            a[v]=node(y1,y2,x1,,V[r]);
            hh[++v]=y2;
            a[v]=node(y1,y2,x2,-,V[r]);
        }
        sort(hh+,hh++v);
        sort(a+,a++v);
        int d=;
        for(i=;i<=v;i++)
            if(hh[i]!=hh[i-])
                hh[++d]=hh[i];
        double ans=;
        V[]=;
        sort(V,V+m+);
        int ct =;
        for(j=;j<=m;j++)
        {
            ct=;
            for(i=;i<=v;i++)
                if(a[i].v>V[j-])
                    tmp[ct++]=a[i];
            mem(col,);
            mem(len,);
            for(i=;i<ct-;i++)
            {
                //int l=BinarySearch(tmp[i].l,1,M);
                //int r=BinarySearch(tmp[i].r,1,M)-1;
                int l = lower_bound(hh+,hh+d,tmp[i].l)-hh;
                int r = lower_bound(hh+,hh+d,tmp[i].r)-hh-;
                if(l<=r) update(tmp[i].c,l,r,,d,);
                ans+=len[]*(double)(V[j]-V[j-])*(tmp[i+].x-tmp[i].x);
            }
        }
        pf("Case %d: %.0lf\n",kase++,ans);
    }
    return ;
}