Pinball Game 3D
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 688 Accepted Submission(s): 276
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.
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.
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 3
2 1 3 2HintIn the first case, RD can shoot the second ball at first and hit the third ball indirectly. In the second case, RD can shoot the second or the third ball initially and hit the fourth ball as well as the fifth ball. Two schemes are both the best.
#include<algorithm>
#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=100010;
const int mod=1<<30;
typedef long long ll;
struct node
{
int x,y,z;
inline bool operator <(const node& tt) const
{
if(x!=tt.x)
return x<tt.x;
if(y!=tt.y)
return y<tt.y;
return z<tt.z;
}
} po[maxn];
int dp[maxn],sy[maxn],sz[maxn],mv[maxn],ans;
int tmp[maxn<<1],ty[maxn],ptr;
ll num[maxn],nv[maxn],ansn;
bool cmp(int a,int b)//注意y必须这么排.不然可能出现y相等的时候i+1排到i前面的情况.
{
if(po[a].y!=po[b].y)
return po[a].y<po[b].y;
return a<b;
}
void update(int x,int v,ll d,int mn)
{
for(int i=x;i<=mn;i+=i&-i)
{
if(v>mv[i])
mv[i]=v,nv[i]=d;
else if(v==mv[i])
nv[i]+=d;
}
}
int qu(int x,ll &nn)
{
int ret=0;
for(int i=x;i>0;i-=i&-i)
{
if(ret<mv[i])
ret=mv[i],nn=nv[i];
else if(mv[i]==ret)
nn+=nv[i];
}
return ret;
}
void solve(int l,int r)
{
if(l==r)
{
if(dp[l]>ans)
ans=dp[l],ansn=num[l];
else if(dp[l]==ans)
ansn+=num[l];
sz[l]=po[l].z;
return;
}
int mid=(l+r)>>1,le=l,ri=mid+1,len=r-l+1,ml=mid-l+1,lim,h,i;
memmove(tmp+ptr,sy+l,len*sizeof(int));
for(i=0;i<len;i++)
if(tmp[ptr+i]<=mid)//直接把排好的序划分下去就行了。
sy[le++]=tmp[ptr+i];
else
sy[ri++]=tmp[ptr+i];
ptr+=len;
solve(l,mid);
lim=ptr,ptr-=len;
for(i=1;i<=ml;i++)
mv[i]=0;
for(i=ptr;i<lim;i++)
{
if(tmp[i]<=mid)
{
h=lower_bound(sz+l,sz+l+ml,po[tmp[i]].z)-(sz+l)+1;
update(h,dp[tmp[i]],num[tmp[i]],ml);
continue;
}
h=lower_bound(sz+l,sz+l+ml,po[tmp[i]].z)-(sz+l);
if(h>=ml||sz[l+h]>po[tmp[i]].z)
h--;
if(h>=0)
{
ll cc=0;
int tt=qu(h+1,cc)+1;
if(tt>dp[tmp[i]])
dp[tmp[i]]=tt,num[tmp[i]]=cc;
else if(tt==dp[tmp[i]]&&tt!=1)
num[tmp[i]]+=cc;
}
}
solve(mid+1,r);
merge(sz+l,sz+l+ml,sz+l+ml,sz+l+len,ty);
memmove(sz+l,ty,len*sizeof(int));
}
int main()
{
int t,n,i;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(i=1;i<=n;i++)
scanf("%d%d%d",&po[i].x,&po[i].y,&po[i].z);
ansn=ptr=ans=0;
sort(po+1,po+n+1);
for(i=1;i<=n;i++)
sy[i]=i,dp[i]=1,num[i]=1;
sort(sy+1,sy+n+1,cmp);
solve(1,n);
printf("%d %I64d\n",ans,ansn%mod);
}
return 0;
}
/*
送一组数据。网上好多代码都过不了这组。就是因为排序y的时候i+1排到i前面去了。导致少更新很多
杭电数据比较水就水过去了。
1
17
2 0 0
2 1 1
1 2 1
1 0 0
0 0 1
0 2 0
0 2 2
0 1 1
2 0 1
1 2 2
0 1 0
2 0 2
2 2 2
0 1 2
1 0 2
1 1 1
0 0 2
*/