HDU 3642 Get The Treasury (线段树体积交[转成面积交])

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#线段树

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本文介绍了一种求解多个长方体在三维空间中至少三次重叠的体积交集的方法。通过将问题转化为二维平面上的面积交集计算，并采用线段树等数据结构优化计算过程。

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题意：

求n 个长方体的体积交（至少三次以上）。

思路：

因为Z的范围比较小。

直接暴力Z轴，把所有Z轴转成二维的平面。

直接求z个面积交即可。

注意区间是一闭一开。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;


const int maxn = 2000 + 10;
int T, ks;

struct node{
    int l,r;
    int len[4];
    int addr;
}nod[maxn<<2];


struct RECT{
    int x,y,x1,y1;
    RECT(int x = 0,int y = 0,int x1 = 0, int y1 = 0):x(x), y(y), x1(x1), y1(y1){}
};

vector<RECT>v[maxn>>1];

struct LINE{
    int l,r;
    int h;
    int cnt;
    LINE(int l = 0,int r = 0,int h = 0,int cnt = 0):l(l), r(r), h(h), cnt(cnt){}
}line[maxn];
bool cmp(const LINE& lhs, const LINE& rhs){
    return lhs.h < rhs.h || (lhs.h == rhs.h && lhs.cnt > rhs.cnt);
}
int a[maxn];
int cnt, id2;

int get(int x){
    int l = 0,r = id2 - 1, m;
    while(l <= r){
        m = l + r >> 1;
        if (a[m] == x) return m;
        else if (a[m] > x){
            r = m - 1;
        }
        else l = m + 1;
    }
}
void pushup(int o){
    int l = nod[o].l;
    int r = nod[o].r;

    if (l == r){
        if (nod[o].addr >= 3){
            nod[o].len[3] = nod[o].len[0];
            nod[o].len[2] = nod[o].len[1] = 0;
        }
        else if (nod[o].addr >= 2){
            nod[o].len[3] = nod[o].len[1] = 0;
            nod[o].len[2] = nod[o].len[0];
        }
        else if (nod[o].addr >= 1){
            nod[o].len[2] = nod[o].len[3] = 0;
            nod[o].len[1] = nod[o].len[0];
        }
        else {
            nod[o].len[1] = nod[o].len[2] = nod[o].len[3] = 0;
        }
        return;
    }

    int lson = o << 1;
    int rson = o << 1 | 1;

    if (nod[o].addr >= 3){
        nod[o].len[3] = nod[o].len[0];
        nod[o].len[1] = nod[o].len[2] = 0;
    }
    else if (nod[o].addr >= 2){
        nod[o].len[3] = 0;
        for (int i = 1; i <= 3; ++i){
            nod[o].len[3] += nod[lson].len[i];
            nod[o].len[3] += nod[rson].len[i];
        }
        nod[o].len[1] = 0;
        nod[o].len[2] = nod[o].len[0] - nod[o].len[3];
    }
    else if (nod[o].addr >= 1){
        nod[o].len[3] = 0;
        for (int i = 2; i <= 3; ++i){
            nod[o].len[3] += nod[lson].len[i];
            nod[o].len[3] += nod[rson].len[i];
        }
        nod[o].len[2] = nod[lson].len[1] + nod[rson].len[1];
        nod[o].len[1] = nod[o].len[0] - nod[o].len[3] - nod[o].len[2];
    }
    else {
        nod[o].len[3] = nod[lson].len[3] + nod[rson].len[3];
        nod[o].len[2] = nod[lson].len[2] + nod[rson].len[2];
        nod[o].len[1] = nod[lson].len[1] + nod[rson].len[1];
    }
}
void build(int l,int r,int o){
    nod[o].l = l;
    nod[o].r = r;
    nod[o].addr = 0;
    nod[o].len[0] = a[r] - a[l-1];
    for (int i = 1; i <= 3; ++i){
        nod[o].len[i] = 0;
    }
    if (l == r) {
        return;
    }
    int m = l + r >> 1;
    int lson = o << 1;
    int rson = o << 1 | 1;
    build(l,m,lson);
    build(m+1,r,rson);
    nod[o].len[0] = nod[lson].len[0] + nod[rson].len[0];
}

void update(int L,int R,int c,int l,int r,int o){
    if (L <= l && r <= R){
        nod[o].addr += c;
        pushup(o);
        return;
    }

    int m = l + r >> 1;
    if (m >= L){
        update(L,R,c,l,m,o<<1);
    }
    if (m < R){
        update(L,R,c,m+1,r,o<<1|1);
    }
    pushup(o);
}


long long solve(int id){
    if ((int)v[id].size() < 3) return 0;
    cnt = id2 = 0;
    for (int i = 0; i < v[id].size(); ++i){
        int x = v[id][i].x;
        int y = v[id][i].y;
        int x1 = v[id][i].x1;
        int y1 = v[id][i].y1;
        line[cnt++] = LINE(x,x1, y, 1);
        line[cnt++] = LINE(x,x1, y1, -1);
        a[id2++] = x;
        a[id2++] = x1;
    }

    sort(a,a+id2);
    id2 = unique(a, a + id2) - a;
    sort(line, line + cnt, cmp);
    build(1, id2-1, 1);
    update(get(line[0].l) + 1, get(line[0].r), line[0].cnt, 1, id2 - 1, 1);

    long long ans = 0;
    for (int i = 1; i < cnt; ++i){
        ans += (long long)nod[1].len[3] * (line[i].h - line[i-1].h);
        update(get(line[i].l) + 1, get(line[i].r), line[i].cnt, 1, id2 - 1, 1);
    }
    return ans;
}

int main(){
    scanf("%d",&T);
    while(T--){
        for (int i = 0; i < (maxn >> 1); ++i ){
            v[i].clear();
        }

        int n;
        scanf("%d",&n);
        for (int i = 0; i < n; ++i){
            int x,y,z,x1,y1,z1;
            scanf("%d %d %d %d %d %d", &x, &y, &z, &x1, &y1, &z1);
            for (int j = z; j < z1; ++j){
                v[j + 500].push_back(RECT(x,y,x1,y1));
            }
        }
        long long ans = 0;
        for (int i = 0; i <= 1000; ++i){
            ans += solve(i);
        }
        printf("Case %d: %I64d\n", ++ks, ans);
    }



    return 0;
}

Get The Treasury

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2720 Accepted Submission(s): 881

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

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HDU 3642 Get The Treasury (线段树 体积交[转成面积交])

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HDU 3642 Get The Treasury (线段树体积交[转成面积交])