HDU 4247 Pinball Game 3D(cdq 分治+树状数组+动态规划)
HDU 4247 Pinball Game 3D(cdq 分治+树状数组+动态规划)
Pinball Game 3D
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1137 Accepted Submission(s): 477
Problem Description
RD is a smart boy and excel in pinball game. However, playing common 2D pinball game for a great number of times results in accumulating tedium. Recently, RD has found a new type of pinball game, a 3D pinball game. The 3D pinball game space can be regarded as a three dimensional coordinate system containing N balls. A ball can be considered as a point. At the beginning, RD made a shot and hit a ball. The ball hit by RD will move and may hit another ball and the “another ball” may move and hit another another ball, etc. But once a ball hit another ball, it will disappear. RD is skilled in this kind of game, so he is able to control every ball's moving direction. But there is a limitation: if ball A's coordinate is (x1,y1,z1) and ball B's coordinate is (x2,y2,z2), then A can hit B only if x1 <= x2 and y1 <= y2 and z1 <= z2. Now, you should help RD to calculate the maximum number of balls that can be hit and the number of different shooting schemes that can achieve that number. Two schemes are different if the sets of hit balls are not the same. The order doesn't matter.
Input
The first line contains one integer T indicating the number of cases. In each case, the first line contains one integer N indicating the number of balls. The next N lines each contains three non-negative integer (x, y, z), indicating the coordinate of a ball. The data satisfies T <= 3, N <= 105, 0 <= x, y, z <= 230, no two balls have the same coordinate in one case.
Output
Print two integers for each case in a line, indicating the maximum number of balls that can be hit and the number of different shooting schemes. As the number of schemes can be quite large, you should output this number mod 230.
Sample Input
2
3
2 0 0
0 1 0
0 1 1
5
3 0 0
0 1 0
0 0 1
0 2 2
3 3 3Sample Output
2 1
3 2
问题是求三维的LIS问题。LIS，即最长递增子序列。显然是用动态规划来求解的一维的情况，可以直接两个for循环，进行DP。但是如果数据有1e5,暴力循环效率是O(n^2*(n+1)/2)，所以可以用一些数据结构优化一下，比如单调队列，树状数组等。树状数组最擅长的是快速的求前缀和，同时也可以求前缀和最值。下面三维的情况要复杂一点。首先，要找到比这个点小的点，然后在这些点中进行DP这就属于三维偏序的问题，对于三维偏序，一般都是降维处理可以一维排序，二维CDQ分治，三维树状数组，这里，CDQ分治，要先处理左半边，然后处理左半边对右半边的影响，再处理右半边。树状数组里面插入DP状态，#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>
using namespace std;
const int maxn=1e5;
const int INF=0x7FFFFFFF;
const int mod = 1 << 30 ;
struct Node
{
int x,y,z;
int id,z2;
}a[maxn+5],b[maxn+5];
int n,e,d[maxn+5];
int cmp(Node a,Node b)
{
if(a.x==b.x&&a.y==b.y)
return a.z<b.z;
else if(a.x==b.x)
return a.y<b.y;
return a.x<b.x;
}
int cmp2(Node a,Node b)
{
return a.z<b.z;
}
struct node
{
int len;
int num;
}dp[maxn+5],c[maxn+5];
int lowbit(int x)
{
return x&(-x);
}
void update(node &term1,node term2)
{
if(term1.len<term2.len)
{
term1=term2;
}
else if(term1.len==term2.len)
term1.num+=term2.num;
}
void insert(int x,node y)
{
for(int i=x;i<=e;i+=lowbit(i))
{
update(c[i],y);
}
}
node sum(int x)
{
node p;
for(int i=x;i>=1;i-=lowbit(i))
{
update(p,c[i]);
}
return p;
}
void del(int x)
{
for(int i=x;i<=e;i+=lowbit(i))
{
c[i].len=0;
c[i].num=0;
}
}
void fun(int l,int r)
{
if(l==r)
{
return;
}
int mid=(l+r)>>1;
fun(l,mid);
for(int i=l;i<=r;i++)
{
b[i]=a[i];
b[i].x=0;
}
sort(b+l,b+r+1,cmp);
for(int i=l;i<=r;i++)
{
if(b[i].id<=mid)
{
insert(b[i].z,dp[b[i].id]);
}
else
{
node temp=sum(b[i].z);
if(dp[b[i].id].len<temp.len+1)
{
dp[b[i].id].len=temp.len+1;
dp[b[i].id].num=temp.num;
}
else if(dp[b[i].id].len==temp.len+1)
dp[b[i].id].num+=temp.num;
}
}
for(int i=l;i<=r;i++)
{
if(b[i].id<=mid)
del(b[i].z);
}
fun(mid+1,r);
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
e=0;
for(int i=1;i<=n;i++)
{
scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].z);
d[i]=a[i].z;
}
sort(a+1,a+n+1,cmp);
sort(d+1,d+1+n),e=unique(d+1,d+1+n)-d;
for(int i=1;i<=n;i++)
{
a[i].id=i;
dp[i].len=1;
dp[i].num=1;
a[i].z=lower_bound(d+1,d+1+e,a[i].z)-d;
c[i].len=0;
c[i].num=0;
}
fun(1,n);
node ans;
ans.len=0;
ans.num=0;
for(int i=1;i<=n;i++)
update(ans,dp[i]);
printf("%d %d\n",ans.len,ans.num%mod);
}
return 0;
}