HDU3642 Get The Treasury —— 求矩形交体积 线段树 + 扫描线 + 离散化

Posted 2020-10-12 h_z_cong

题目链接：https://vjudge.net/problem/HDU-3642

 

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. 

InputThe 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 z₂, 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. 

OutputFor 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

 

 

 

 

代码如下：

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <cmath>
  5 #include <algorithm>
  6 #include <vector>
  7 #include <queue>
  8 #include <stack>
  9 #include <map>
 10 #include <string>
 11 #include <set>
 12 using namespace std;
 13 typedef long long LL;
 14 const double EPS = 1e-8;
 15 const int INF = 2e9;
 16 const LL LNF = 2e18;
 17 const int MAXN = 1e3+10;
 18 
 19 int times[MAXN<<3];
 20 int one[MAXN<<3], two[MAXN<<3], more[MAXN<<3];
 21 int Z[MAXN<<1], X[MAXN<<1];
 22 
 23 struct Cube
 24 {
 25     int x1, y1, z1,x2, y2, z2;
 26 }cube[MAXN];
 27 
 28 struct Line
 29 {
 30     int le, ri, h, id;
 31     bool operator<(const Line &a)const{
 32         return h<a.h;
 33     }
 34 
 35 }line[MAXN<<1];
 36 
 37 void push_up(int u, int l, int r)
 38 {
 39     if(times[u]>=3)
 40     {
 41         more[u] = X[r] - X[l];
 42         two[u] = X[r] - X[l];
 43         one[u] = X[r] - X[l];
 44     }
 45     else if(times[u]==2)
 46     {
 47         more[u] = (l+1==r)?0:(one[u*2]+one[u*2+1]);
 48         two[u] = X[r] - X[l];
 49         one[u] = X[r] - X[l];
 50     }
 51     else if(times[u]==1)
 52     {
 53         more[u] = (l+1==r)?0:(two[u*2]+two[u*2+1]);
 54         two[u] = (l+1==r)?0:(one[u*2]+one[u*2+1]);
 55         one[u] = X[r] - X[l];
 56     }
 57     else
 58     {
 59         more[u] = (l+1==r)?0:(more[u*2]+more[u*2+1]);
 60         two[u] = (l+1==r)?0:(two[u*2]+two[u*2+1]);
 61         one[u] = (l+1==r)?0:(one[u*2]+one[u*2+1]);
 62     }
 63 }
 64 
 65 void add(int u, int l, int r, int x, int y, int v)
 66 {
 67     if(x<=l && r<=y)
 68     {
 69         times[u] += v;
 70         push_up(u, l, r);
 71         return;
 72     }
 73 
 74     int mid = (l+r)>>1;
 75     if(x<=mid-1) add(u*2, l, mid, x, y, v);
 76     if(y>=mid+1) add(u*2+1, mid, r, x, y, v);
 77     push_up(u, l, r);
 78 }
 79 
 80 int main()
 81 {
 82     int T, n;
 83     scanf("%d", &T);
 84     for(int kase = 1; kase<=T; kase++)
 85     {
 86         scanf("%d", &n);
 87         int numZ = 0;
 88         for(int i = 1; i<=n; i++)
 89         {
 90             scanf("%d%d%d", &cube[i].x1,&cube[i].y1,&cube[i].z1);
 91             scanf("%d%d%d", &cube[i].x2,&cube[i].y2,&cube[i].z2);
 92             Z[++numZ] = cube[i].z1; Z[++numZ] = cube[i].z2;
 93         }
 94 
 95         sort(Z+1, Z+1+numZ);
 96         numZ = unique(Z+1, Z+1+numZ) - (Z+1);
 97 
 98         LL volume = 0;
 99         for(int i = 1; i<numZ; i++)
100         {
101             int numLine = 0, numX = 0;
102             for(int j = 1; j<=n; j++)
103             if(cube[j].z1<=Z[i] && Z[i+1]<=cube[j].z2)
104             {
105                 line[++numLine].le = cube[j].x1; line[numLine].ri = cube[j].x2;
106                 line[numLine].h = cube[j].y1; line[numLine].id = 1;
107                 line[++numLine].le = cube[j].x1; line[numLine].ri = cube[j].x2;
108                 line[numLine].h = cube[j].y2; line[numLine].id = -1;
109                 X[++numX] = cube[j].x1; X[++numX] = cube[j].x2;
110             }
111 
112             sort(line+1, line+1+numLine);
113             sort(X+1, X+1+numX);
114             numX = unique(X+1, X+1+numX) - (X+1);
115 
116             memset(times, 0, sizeof(times));
117             memset(more, 0, sizeof(more));
118             memset(two, 0, sizeof(two));
119             memset(one, 0, sizeof(one));
120 
121             LL area = 0;
122             for(int j = 1; j<numLine; j++)
123             {
124                 int l = upper_bound(X+1, X+1+numX, line[j].le) - (X+1);
125                 int r = upper_bound(X+1, X+1+numX, line[j].ri) - (X+1);
126                 add(1, 1, numX, l, r, line[j].id);
127                 area += 1LL*more[1]*(line[j+1].h-line[j].h);
128             }
129             volume += 1LL*area*(Z[i+1]-Z[i]);
130         }
131         printf("Case %d: %lld\n", kase, volume);
132     }
133 }

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