HDU 3642 Get The Treasury（线段树，求立方体交）
Get The Treasury
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1408 Accepted Submission(s): 440
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.
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.
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
Recommend
lcy
求立方体相交至少3次的体积。
枚举z
然后求矩形面积交。
常规的线段树题目。
较麻烦
/* *HDU 3642 给定一些长方体，求这些长方体相交至少3次的体积 z坐标的绝对值不超过500，所以枚举z坐标，然后求矩形面积并 */ #include <stdio.h> #include <string.h> #include <algorithm> #include <iostream> using namespace std; const int MAXN=2010; struct Node { int l,r; int lf,rf;//实际的左右端点 int c; int once,twice,more; }segTree[MAXN*3]; int y[MAXN]; int z[MAXN]; struct Line { int x; int y1,y2; int z1,z2;//这两个是枚举的时候判断使用的 int f; }line[MAXN]; bool cmp(Line a,Line b) { return a.x<b.x; } void Build(int i,int l,int r) { segTree[i].l=l; segTree[i].r=r; segTree[i].lf=y[l]; segTree[i].rf=y[r]; segTree[i].c=0; segTree[i].once=segTree[i].twice=segTree[i].more=0; if(r==l+1)return; int mid=(l+r)>>1; Build(i<<1,l,mid); Build((i<<1)|1,mid,r); } void push_up(int i) { if(segTree[i].c>2) { segTree[i].more=segTree[i].rf-segTree[i].lf; segTree[i].once=segTree[i].twice=0; } else if(segTree[i].c==2) { if(segTree[i].l+1==segTree[i].r)//叶子结点 { segTree[i].more=0; segTree[i].twice=segTree[i].rf-segTree[i].lf; segTree[i].once=0; return; } segTree[i].more=segTree[i<<1].once+segTree[i<<1].twice+segTree[i<<1].more +segTree[(i<<1)|1].once+segTree[(i<<1)|1].twice+segTree[(i<<1)|1].more; segTree[i].twice=segTree[i].rf-segTree[i].lf-segTree[i].more; segTree[i].once=0; } else if(segTree[i].c==1) { if(segTree[i].l+1==segTree[i].r) { segTree[i].more=0; segTree[i].twice=0; segTree[i].once=segTree[i].rf-segTree[i].lf; return; } segTree[i].more=segTree[i<<1].more+segTree[i<<1].twice +segTree[(i<<1)|1].more+segTree[(i<<1)|1].twice; segTree[i].twice=segTree[i<<1].once+segTree[(i<<1)|1].once; segTree[i].once=segTree[i].rf-segTree[i].lf-segTree[i].more-segTree[i].twice; } else { if(segTree[i].l+1==segTree[i].r) { segTree[i].more=segTree[i].once=segTree[i].twice=0; return; } segTree[i].more=segTree[i<<1].more+segTree[(i<<1)|1].more; segTree[i].twice=segTree[i<<1].twice+segTree[(i<<1)|1].twice; segTree[i].once=segTree[i<<1].once+segTree[(i<<1)|1].once; } } void update(int i,Line e) { if(segTree[i].lf>=e.y1 && segTree[i].rf<=e.y2) { segTree[i].c+=e.f; push_up(i); return; } if(e.y2<=segTree[i<<1].rf) update(i<<1,e); else if(e.y1>=segTree[(i<<1)|1].lf) update((i<<1)|1,e); else { Line temp=e; temp.y2=segTree[i<<1].rf; update(i<<1,temp); temp=e; temp.y1=segTree[(i<<1)|1].lf; update((i<<1)|1,temp); } push_up(i); } Line temp[MAXN]; int main() { int T; int n; int x1,y1,z1,x2,y2,z2; scanf("%d",&T); int iCase=0; while(T--) { iCase++; scanf("%d",&n); int t=0; for(int i=0;i<n;i++) { scanf("%d%d%d%d%d%d",&x1,&y1,&z1,&x2,&y2,&z2); line[t].x=x1; line[t].y1=y1; line[t].y2=y2; line[t].z1=z1; line[t].z2=z2; line[t].f=1; y[t]=y1; z[t++]=z1; line[t].x=x2; line[t].y1=y1; line[t].y2=y2; line[t].z1=z1; line[t].z2=z2; line[t].f=-1; y[t]=y2; z[t++]=z2; } sort(line,line+t,cmp); sort(y,y+t); int t1=unique(y,y+t)-y; Build(1,0,t1-1); sort(z,z+t); int t2=unique(z,z+t)-z; long long ans=0; long long area=0; for(int i=0;i<t2-1;i++) { int m=0; for(int j=0;j<t;j++) if(line[j].z1<=z[i]&&line[j].z2>z[i]) temp[m++]=line[j]; area=0; update(1,temp[0]); for(int j=1;j<m;j++) { area+=(long long)segTree[1].more*(temp[j].x-temp[j-1].x); update(1,temp[j]); } ans+=area*(z[i+1]-z[i]); } printf("Case %d: %I64d\n",iCase,ans); } return 0; }