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Get The Treasury

题目链接

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

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 x1, y1, z1, x2, y2 and z2 (x1<x2, y1<y2, z1<z2). 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 x1 to x2. 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 x1, y1, z1, x2, y2 and z2, separated by a space. The absolute value of x and y coordinates of the vertices is no more than 106, 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

题意

给你一些立方体，求相交三次或以上的体积和。

题解

昨天做了个求面积交两次或以上的，今天升级了，交三次其实还好，和交两次超不多，但是球体积我就有点懵了，想了好久，不知道咋办。

最终看题解，结果。。。感觉这不是暴力吗？多出来的一维z，先把z排序，然后每次把包含z[i]到z[i+1]的立方体加入到一个零食队列中，求这些立方体的面积并，在乘上(z[i+1]-z[i])。

既然这样可以过，那其实和二维没什么差别了，无非就是代码多了一些，都是套路啊！

代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define INF 0x7f7f7f7f
#define N 1050
ll kthx[N<<1],kthz[N<<1];
template<typename T>void read(T&x)
{
    ll k=0; char c=getchar();
    x=0;
    while(!isdigit(c)&&c!=EOF)k^=c=='-',c=getchar();
    if (c==EOF)exit(0);
    while(isdigit(c))x=x*10+c-'0',c=getchar();
    x=k?-x:x;
}
struct Query
{
    ll l,r,z1,z2,h; int id;
    bool operator<(const Query&e)const
        {return h<e.h;}
}que[N<<1],tmp[N<<1];
struct Node{int l,r,lazy;ll sum,sum2,sum3;};
struct segmentTree
{
    Node tr[N<<3];
    void push_up(int x);
    void bt(int x,int l,int r);
    void update(int x,int l,int r,int tt);
}seg;
void segmentTree::push_up(int x)
{
    ll len=kthx[tr[x].r+1]-kthx[tr[x].l];
    tr[x].sum=0;
    if (tr[x].l<tr[x].r)tr[x].sum=tr[x<<1].sum+tr[x<<1|1].sum;
    if (tr[x].lazy>=1)tr[x].sum=len;
    tr[x].sum2=0;
    if (tr[x].lazy>=2)tr[x].sum2=len;
    if (tr[x].l<tr[x].r&&tr[x].lazy==1)tr[x].sum2=tr[x<<1].sum+tr[x<<1|1].sum;
    if (tr[x].l<tr[x].r&&tr[x].lazy==0)tr[x].sum2=tr[x<<1].sum2+tr[x<<1|1].sum2;
    tr[x].sum3=0;
    if (tr[x].lazy>=3)tr[x].sum3=len;
    if (tr[x].l==tr[x].r)return;
    if (tr[x].lazy==2)tr[x].sum3=tr[x<<1].sum+tr[x<<1|1].sum;
    if (tr[x].lazy==1)tr[x].sum3=tr[x<<1].sum2+tr[x<<1|1].sum2;
    if (tr[x].lazy==0)tr[x].sum3=tr[x<<1].sum3+tr[x<<1|1].sum3;
}
void segmentTree::bt(int x,int l,int r)
{
    tr[x]=Node{l,r,0,0,0,0};
    if(l==r)return;
    int mid=(l+r)>>1;
    bt(x<<1,l,mid);
    bt(x<<1|1,mid+1,r);
}
void segmentTree::update(int x,int l,int r,int tt)
{
    if (l<=tr[x].l&&tr[x].r<=r)
    {
        tr[x].lazy+=tt;
        push_up(x);
        return;
    }
    int mid=(tr[x].l+tr[x].r)>>1;
    if(l<=mid)update(x<<1,l,r,tt);
    if(mid<r)update(x<<1|1,l,r,tt);
    push_up(x);
}
void work()
{
    int m,numx=0,numz=0;
    ll ans=0;
    read(m);
    for(int i=1;i<=m;i++)
    {
        ll x1,y1,z1,x2,y2,z2;
        //scanf("%lld%lld%lld%lld%lld%lld",&x1,&y1,&z1,&x2,&y2,&z2);
        read(x1); read(y1); read(z1); read(x2); read(y2); read(z2);
        que[i]={x1,x2,z1,z2,y1,1};
        que[i+m]={x1,x2,z1,z2,y2,-1};
        kthx[++numx]=x1;
        kthx[++numx]=x2;
        kthz[++numz]=z1;
        kthz[++numz]=z2;
    }
    sort(que+1,que+numx+1);
    sort(kthx+1,kthx+numx+1);
    sort(kthz+1,kthz+numz+1);
    numx=unique(kthx+1,kthx+numx+1)-kthx-1;
    numz=unique(kthz+1,kthz+numz+1)-kthz-1;
    for(int i=1;i<=2*m;i++)
        {
            que[i].l=lower_bound(kthx+1,kthx+numx+1,que[i].l)-kthx;
            que[i].r=lower_bound(kthx+1,kthx+numx+1,que[i].r)-kthx-1;
        }
    seg.bt(1,1,numx);
    for(int j=1;j<=numz-1;j++)
    {
        ll z1=kthz[j],z2=kthz[j+1],tp=0,now=0;
        for(int i=1;i<=2*m;i++)
            if (que[i].z1<=z1&&z2<=que[i].z2)tmp[++now]=que[i];
        for(int i=1;i<=now;i++)
            {
                seg.update(1,tmp[i].l,tmp[i].r,tmp[i].id);
                tp+=seg.tr[1].sum3*(tmp[i+1].h-tmp[i].h); 
            }
        ans+=1LL*tp*(z2-z1);
    }
    static int cas;
    printf("Case %d: %lld\n",++cas,ans);
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("aa.in","r",stdin);
#endif
    int T;
    read(T);
    while(T--)work();
}

HDU 3642 求体积交集

标签：nes   range   rate   long   operator   surface   http   href   一个   

原文地址：https://www.cnblogs.com/mmmqqdd/p/11247711.html