HDU 3255 Farming(线段树求体积并)

http://acm.hdu.edu.cn/showproblem.php?pid=3255

FarmingTime Limit: 12000/6000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1226    Accepted Submission(s): 354

Problem Description

You have a big farm, and you want to grow vegetables in it. You're too lazy to seed the seeds yourself, so you've hired n people to do the job for you.
Each person works in a rectangular piece of land, seeding one seed in one unit square. The working areas of different people may overlap, so one unit square can be seeded several times. However, due to limited space, different seeds in one square fight each other -- finally, the most powerful seed wins. If there are several "most powerful" seeds, one of them win (it does not matter which one wins).

There are m kinds of seeds. Different seeds grow up into different vegetables and sells for different prices. 
As a rule, more powerful seeds always grow up into more expensive vegetables.
Your task is to calculate how much money will you get, by selling all the vegetables in the whole farm.

 

Input

The first line contains a single integer T (T <= 10), the number of test cases. 
Each case begins with two integers n, m (1 <= n <= 30000, 1 <= m <= 3).
The next line contains m distinct positive integers p_i (1 <= p_i <= 100), the prices of each kind of vegetable. 
The vegetables (and their corresponding seeds) are numbered 1 to m in the order they appear in the input. 
Each of the following n lines contains five integers x1, y1, x2, y2, s, indicating a working seeded a rectangular area with lower-left corner (x1,y1), upper-right corner (x2,y2), with the s-th kind of seed.
All of x1, y1, x2, y2 will be no larger than 10⁶ in their absolute values.

 

Output

For each test case, print the case number and your final income.

 

Sample Input

21 1250 0 10 10 12 25 20 0 2 1 11 0 3 2 2

 

Sample Output

Case 1: 2500Case 2: 16

 

Source

2009 Asia Regional Ningbo Online

 

你家种一年地的收成暴int！你当是《日人民报》啊！

题意：

雇佣贫农种地（所以是地主喽），每个农民带一种种子，播种在他管理矩形区域，每个单位面积种一粒，区域可能会重合，那么收成高的那粒会将其他的覆盖掉，求最终收成。

分析：

一开始想的是二维线段树set,但是看到面积有sqr(10^6),即使离散化也有sqr(3*10^4),后来想到加权的线段树扫描线求面积并，但是发现覆盖种子不好维护，后来看到别人题解标题：线段树求体积并。

把价格看成高度，那么每粒种子的高度区间就可以看成[0,price),想像一下，这些"长方体"的形状一定是"类金字塔"形的，也就是空间中每一个微元一定是"接地的"，那么我们只要对价格排序，然后用类似微积分的思想（不会的下学期自觉重修去）求出各部分的体积（最多3次求面积并，面积×高度/价格差）即可。

/* * *Author:fcbruce * *Date:2014-08-27 14:14:21  * */#include <cstdio>#include <iostream>#include <sstream>#include <cstdlib>#include <algorithm>#include <ctime>#include <cctype>#include <cmath>#include <string>#include <cstring>#include <stack>#include <queue>#include <list>#include <vector>#include <map>#include <set>#define sqr(x) ((x)*(x))#define LL long long#define itn int#define INF 0x3f3f3f3f#define PI 3.1415926535897932384626#define eps 1e-10#ifdef _WIN32#define lld "%I64d"#else#define lld "%lld"#endif#define maxm #define maxn 30303using namespace std;struct __seed{int x1,y1,x2,y2,price;}seed[maxn];struct __edge{int x1,x2,y,type;bool operator < (const __edge &e)const{return y<e.y;}}edge[maxn<<1];int price[4];int X[maxn<<1];int add[maxn<<3],sum[maxn<<3];inline void pushup(int k,int l,int r){if (add[k]>0)sum[k]=X[r]-X[l];else if (r-l==1)sum[k]=0;elsesum[k]=sum[k*2+1]+sum[k*2+2];}void update(int a,int b,int v,itn k,int l,int r){if (b<=l || r<=a)return ;if (a<=l && r<=b){add[k]+=v;}else{update(a,b,v,k*2+1,l,l+r>>1);update(a,b,v,k*2+2,l+r>>1,r);}pushup(k,l,r);}int main(){#ifndef ONLINE_JUDGEfreopen("/home/fcbruce/文档/code/t","r",stdin);#endif // ONLINE_JUDGEint n,m,T_T;scanf( "%d",&T_T);for (int __=1;__<=T_T;__++){scanf( "%d%d",&n,&m);price[0]=0;for (int i=1;i<=m;i++)scanf( "%d",price+i);for (int i=0,x1,y1,x2,y2,type;i<n;i++){scanf( "%d%d%d%d%d",&x1,&y1,&x2,&y2,&type);seed[i]=(__seed){x1,y1,x2,y2,price[type]};}sort(price+1,price+m+1);LL income=0;for (int i=1;i<=m;i++){int cnt=0;for (int j=0;j<n;j++){if (seed[j].price>=price[i]){edge[cnt]=(__edge){seed[j].x1,seed[j].x2,seed[j].y1,1};X[cnt++]=seed[j].x1;edge[cnt]=(__edge){seed[j].x1,seed[j].x2,seed[j].y2,-1};X[cnt++]=seed[j].x2;}}sort(edge,edge+cnt);sort(X,X+cnt);int xn=unique(X,X+cnt)-X;memset(sum,0,sizeof sum);memset(add,0,sizeof add);LL area=0;for (int j=0;j<cnt;j++){if (j) area+=(LL)(edge[j].y-edge[j-1].y)*sum[0];int a=lower_bound(X,X+xn,edge[j].x1)-X;int b=lower_bound(X,X+xn,edge[j].x2)-X;update(a,b,edge[j].type,0,0,xn+1);}income+=(LL)(price[i]-price[i-1])*area;}printf( "Case %d: ",__);printf( lld "\n",income);}return 0;}