hdu 3255 Farming(线段树求体积并)

Farming

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1445    Accepted Submission(s): 428

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 10 ⁶ in their absolute values.

 

Output

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

 

Sample Input

2 1 1 25 0 0 10 10 1 2 2 5 2 0 0 2 1 1 1 0 3 2 2

 

Sample Output

Case 1: 2500 Case 2: 16

 

Source

2009 Asia Regional Ningbo Online

 

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题意：

原始题意就不说了。抽象出来就是给你n个底面在xoy平面与坐标轴平行的立方体。他们之间可以重叠。问体积并是多少。

思路：

和求面积并的思路类似。用高度（价值）把立方体分成很多段。然后一段一段求体积就好。

详细见代码：

#include
#include
#include
using namespace std;
const int maxn=30010;
typedef long long ll;
#define lson L,mid,ls
#define rson mid+1,R,rs
struct node
{
    int x1,x2,h,tag;//面积并用的值。tag表示加边还是减边
    int p;//存第三维高度。
    node(){}
    node(int a,int b,int c,int d,int e):x1(a),x2(b),h(c),tag(d),p(e){}
    bool operator <(const node &op) const
    {
        return h>1;
    build(lson);
    build(rson);
}
void update(int L,int R,int rt,int l,int r,int d)
{
    if(l<=L&&R<=r)
    {
        cov[rt]+=d;
        len[rt]=cov[rt]?H[R+1]-H[L]:(L==R?0:len[rt<<1]+len[rt<<1|1]);
        //printf("%d -> %d len %d\n",H[L],H[R+1],len[rt]);
        return;
    }
    int ls=rt<<1,rs=ls|1,mid=(L+R)>>1;
    if(l<=mid)
        update(lson,l,r,d);
    if(r>mid)
        update(rson,l,r,d);
    len[rt]=cov[rt]?H[R+1]-H[L]:len[ls]+len[rs];
    //printf("%d -> %d len %d\n",H[L],H[R+1],len[rt]);
}
int main()
{
    int t,n,m,i,j,cas=1,ct,cnt;
    int x1,x2,y1,y2,ch,ns;
    ll ans,tp;

    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(i=1;i<=m;i++)
            scanf("%d",&val[i]);
        ans=ns=ct=0;
        for(i=1;i<=n;i++)
        {
            scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&ch);
            seg[ns++]=node(x1,x2,y1,1,val[ch]);
            seg[ns++]=node(x1,x2,y2,-1,val[ch]);
            H[ct++]=x1,H[ct++]=x2;
        }
        sort(H,H+ct);
        ct=unique(H,H+ct)-H;
        sort(seg,seg+ns);
        sort(val+1,val+m+1);
        for(i=0;ival[i])
                    tmp[cnt++]=seg[j];
            build(0,ct-2,1);
            for(j=tp=0,cnt--;j