Farming - HDU 3255 扫描线

Farming

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1307    Accepted Submission(s): 385

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

 

题意：在一个田地上会种植不用种类的作物，如果一个地方多次种植，会长出价值高的那种，问最后的收益是多少。

思路：将作物的价值看成是高，这样的话，就转换成了求长方体的体积。

AC代码如下：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long ll;struct node1{    int l,r,lazy,sum[2];}tree[8000010];struct node2{    int l,r,h,f;}line[60100];struct node3{    int x1,x2,y1,y2,p;}area[30010];bool cmp(node2 a,node2 b){    return a.h<b.h;}int val[10];ll ans;int px[60010],point[2000010];void build(int o,int l,int r){    tree[o].l=l;    tree[o].r=r;    tree[o].lazy=0;    tree[o].sum[1]=0;tree[o].sum[0]=px[r]-px[l-1];    if(l==r)      return;    int mi=(l+r)/2;    build(o<<1,l,mi);    build(o<<1|1,mi+1,r);;}void solve(int o){    if(tree[o].lazy>0)      tree[o].sum[1]=tree[o].sum[0];    else    {        if(tree[o].l==tree[o].r)          tree[o].sum[1]=0;        else          tree[o].sum[1]=tree[o<<1].sum[1]+tree[o<<1|1].sum[1];    }}void update(int o,int l,int r,int f){    if(tree[o].l==l && tree[o].r==r && tree[o].lazy+f>=0)    {        tree[o].lazy+=f;        solve(o);        return;    }    int mi=(tree[o].l+tree[o].r)/2;    if(r<=mi)      update(o<<1,l,r,f);    else if(l>mi)      update(o<<1|1,l,r,f);    else    {        update(o<<1,l,mi,f);        update(o<<1|1,mi+1,r,f);    }    solve(o);}int main(){    int T,n,m,t,i,j,k,num,len;    scanf("%d",&T);    for(t=1;t<=T;t++)    {        ans=0;        scanf("%d%d",&n,&m);        for(i=1;i<=m;i++)           scanf("%d",&val[i]);        for(i=1;i<=n;i++)        {            scanf("%d%d%d%d%d",&area[i].x1,&area[i].y1,&area[i].x2,&area[i].y2,&area[i].p);            area[i].x1+=1000000;            area[i].x2+=1000000;            area[i].p=val[area[i].p];        }        len=-1;        for(i=1;i<=n;i++)        {            px[i*2-1]=area[i].x1;            px[i*2]=area[i].x2;        }        sort(px+1,px+1+n*2);        px[0]=-1;        for(i=1;i<=n*2;i++)           if(px[i]!=px[i-1])           {               px[++len]=px[i];               point[px[i]]=len;           }        for(i=1;i<=n;i++)        {            area[i].x1=point[area[i].x1];            area[i].x2=point[area[i].x2];        }        sort(val+1,val+1+m);        for(k=1;k<=m;k++)        {            num=0;            for(i=1;i<=n;i++)               if(area[i].p>=val[k])               {                   num+=2;                   line[num-1].l=area[i].x1+1;line[num-1].r=area[i].x2;line[num-1].h=area[i].y2;line[num-1].f=-1;                   line[num].l=area[i].x1+1;line[num].r=area[i].x2;line[num].h=area[i].y1;line[num].f=1;               }            sort(line+1,line+1+num,cmp);            build(1,1,len);            for(i=1;i<num;i++)            {                update(1,line[i].l,line[i].r,line[i].f);                ans+=(ll)(val[k]-val[k-1])*tree[1].sum[1]*(line[i+1].h-line[i].h);            }        }        printf("Case %d: %I64d\n",t,ans);    }}