hdu 3642 Get The Treasury

Get The Treasury

http://acm.hdu.edu.cn/showproblem.php?pid=3642

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description Jack knows that there is a great underground treasury in a secret region. And he has a special device that can be used to detect treasury under the surface of the earth. One day he got outside with the device to ascertain the treasury. He chose many different locations on the surface of the earth near the secret region. And at each spot he used the device to detect treasury and got some data from it representing a region, which may contain treasury below the surface. The data from the device at each spot is six integers x 1, y 1, z 1, x 2, y 2 and z 2 (x 1<x 2, y 1<y 2, z 1<z 2). According to the instruction of the device they represent the range of x, y and z coordinates of the region. That is to say, the x coordinate of the region, which may contain treasury, ranges from x 1 to x 2. So do y and z coordinates. The origin of the coordinates is a fixed point under the ground.

Jack can’t get the total volume of the treasury because these regions don’t always contain treasury. Through years of experience, he discovers that if a region is detected that may have treasury at more than two different spots, the region really exist treasure. And now Jack only wants to know the minimum volume of the treasury.

Now Jack entrusts the problem to you.

Input The first line of the input file contains a single integer t, the number of test cases, followed by the input data for each test case.

Each test case is given in some lines. In the first line there is an integer n (1 ≤ n ≤ 1000), the number of spots on the surface of the earth that he had detected. Then n lines follow, every line contains six integers x 1, y 1, z 1, x 2, y 2 and z 2, separated by a space. The absolute value of x and y coordinates of the vertices is no more than 10 6, and that of z coordinate is no more than 500.

Output For each test case, you should output “Case a: b” in a single line. a is the case number, and b is the minimum volume of treasury. The case number is counted from one.  

Sample Input 2 1 0 0 0 5 6 4 3 0 0 0 5 5 5 3 3 3 9 10 11 3 3 3 13 20 45  

Sample Output Case 1: 0 Case 2: 8  

Source 2010 Asia Regional Hangzhou Site —— Online Contest  

Recommend lcy   |   We have carefully selected several similar problems for you:   2871  3308  3641  3397  1540    題意：求n個長方體至少相交3次的體積和 z這一維隻有500，是以枚舉z軸，然後就相當于二維的掃描線 線段樹維護區間完全覆寫1、2、3次的長度  

#include<cstring>
#include<algorithm>
#include<cstdio>
using namespace std;
#define N 1001
#define lc k<<1,l,mid
#define rc k<<1|1,mid+1,r
struct node
{
    int l,r,h,f;
    bool operator < (node p)const
    {
        return h<p.h;
    }
}a[N<<1];
struct edge
{
    int x,xx,y,yy,z,zz;
}b[N];
int sum1[N<<3],sum2[N<<3],sum3[N<<3],f[N<<3],has[N<<1],has2[N<<1];
long long ans;
int n,cnt,opl,opr,w; 
void up(int k,int l,int r)
{
    if(f[k]>=3) sum3[k]=has2[r+1]-has2[l];
    else if(f[k]==2)
    {
        sum3[k]=sum1[k<<1]+sum1[k<<1|1];
        sum2[k]=has2[r+1]-has2[l];
    }
    else if(f[k]==1)
    {
        sum3[k]=sum2[k<<1]+sum2[k<<1|1];
        sum2[k]=sum1[k<<1]+sum1[k<<1|1];
        sum1[k]=has2[r+1]-has2[l];
    }
    else
    {
        sum3[k]=sum3[k<<1]+sum3[k<<1|1];
        sum2[k]=sum2[k<<1]+sum2[k<<1|1];
        sum1[k]=sum1[k<<1]+sum1[k<<1|1];
    }
}
void change(int k,int l,int r)
{
    if(opl<=l && r<=opr)
    {
        f[k]+=w;
        up(k,l,r);
        return;
    }
    int mid=l+r>>1;
    if(opl<=mid) change(lc);
    if(opr>mid) change(rc);
    up(k,l,r); 
}
int main()
{
    int T;
    scanf("%d",&T);
    for(int t=1;t<=T;t++)
    {
        ans=0;
        cnt=0;
        scanf("%d",&n);
        for(int i=1;i<=n;i++) 
        {
            scanf("%d%d%d%d%d%d",&b[i].x,&b[i].y,&b[i].z,&b[i].xx,&b[i].yy,&b[i].zz);
            has[i*2-1]=b[i].z; has[i*2]=b[i].zz;
        }
        sort(has+1,has+2*n+1);
        cnt=unique(has+1,has+2*n+1)-(has+1);
        for(int i=1;i<cnt;i++)
        {
            int sz=0;
            for(int j=1;j<=n;j++)
             if(b[j].z<=has[i] && b[j].zz>=has[i+1])
             {
                 a[++sz].l=b[j].x; a[sz].r=b[j].xx; a[sz].h=b[j].y; a[sz].f=1;
                 a[++sz].l=b[j].x; a[sz].r=b[j].xx; a[sz].h=b[j].yy; a[sz].f=-1;
                 has2[sz-1]=b[j].x; has2[sz]=b[j].xx;
             }
            sort(has2+1,has2+sz+1);
            int m=unique(has2+1,has2+sz+1)-(has2+1);
            sort(a+1,a+sz+1);
            memset(sum1,0,sizeof(sum1));
            memset(sum2,0,sizeof(sum2));
            memset(sum3,0,sizeof(sum3));
            for(int j=1;j<=sz;j++)
            {
                opl=lower_bound(has2+1,has2+m+1,a[j].l)-has2;
                opr=lower_bound(has2+1,has2+m+1,a[j].r)-has2-1;
                w=a[j].f;
                change(1,1,m);
                ans+=1ll*sum3[1]*(a[j+1].h-a[j].h)*(has[i+1]-has[i]);
            }
        }
        printf("Case %d: %I64d\n",t,ans);
    }
}      

轉載于:https://www.cnblogs.com/TheRoadToTheGold/p/7141987.html