Problem D
Prince and Princess
Input: Standard Input
Output: Standard Output
Time Limit: 3 Seconds
In an n x n chessboard, Prince and Princess plays a game. The squares in the chessboard are numbered 1, 2, 3 ... n*n, as shown below:
Prince stands in square 1, make p jumps and finally reach square n*n. He enters a square at most once. So if we use xp to denote the p-th square he enters, then x1, x2, ... xp+1 are all different. Note that x1 = 1 and xp+1 = n*n. Princess does the similar thing - stands in square 1, make q jumps and finally reach square n*n. We use y1, y2 , ... yq+1 to denote the sequence, and all q+1 numbers are different.
Figure 2 belows show a 3x3 square, a possible route for Prince and a different route for Princess.
The Prince moves along the sequence: 1 --> 7 --> 5 --> 4 --> 8 --> 3 --> 9 (Black arrows), while the Princess moves along this sequence: 1 --> 4 --> 3 --> 5 --> 6 --> 2 --> 8 --> 9 (White arrow).
The King -- their father, has just come. "Why move separately? You are brother and sister!" said the King, "Ignore some jumps and make sure that you‘re always together."
For example, if the Prince ignores his 2nd, 3rd, 6th jump, he‘ll follow the route: 1 --> 4 --> 8 --> 9. If the Princess ignores her 3rd, 4th, 5th, 6th jump, she‘ll follow the same route: 1 --> 4 --> 8 --> 9, (The common route is shown in figure 3) thus satisfies the King, shown above. The King wants to know the longest route they can move together, could you tell him?
Input
Output
For each test case, print the case number and the length of longest route. Look at the output for sample input for details.
Sample Input Output for Sample Input
|
1 3 6 7 1 7 5 4 8 3 9 1 4 3 5 6 2 8 9 |
Case 1: 4 |
Problemsetter: Rujia Liu, Member of Elite Problemsetters‘ Panel
Pictures drawn by Shahriar Manzoor, Member of Elite Problemsetters‘ Panel
#include<algorithm>
#include<iostream>
#include<string.h>
#include<sstream>
#include<stdio.h>
#include<math.h>
#include<vector>
#include<string>
#include<queue>
#include<set>
#include<map>
//#pragma comment(linker,"/STACK:1024000000,1024000000")
using namespace std;
const int INF=0x3f3f3f3f;
const double eps=1e-8;
const double PI=acos(-1.0);
const int maxn=63000;
typedef __int64 ll;
int val[maxn<<2],pos[maxn];
int pc[maxn],pcs[maxn];
void btree(int L,int R,int k)
{
int ls,rs,mid;
val[k]=0;
if(L==R)
return ;
ls=k<<1;
rs=ls|1;
mid=(L+R)>>1;
btree(L,mid,ls);
btree(mid+1,R,rs);
}
void update(int L,int R,int p,int k,int v)
{
int ls,rs,mid;
if(L==R)
{
val[k]=v;
return ;
}
ls=k<<1;
rs=ls|1;
mid=(L+R)>>1;
if(p>mid)
update(mid+1,R,p,rs,v);
else
update(L,mid,p,ls,v);
val[k]=max(val[ls],val[rs]);
}
int qu(int L,int R,int l,int r,int k)
{
int ls,rs,mid;
if(l==L&&r==R)
return val[k];
ls=k<<1;
rs=ls|1;
mid=(L+R)>>1;
if(l>mid)
return qu(mid+1,R,l,r,rs);
else if(r<=mid)
return qu(L,mid,l,r,ls);
else
return max(qu(L,mid,l,mid,ls),qu(mid+1,R,mid+1,r,rs));
}
int main()
{
int cas=1,t,n,nc,ns,i,tp,ans;
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d",&n,&nc,&ns);
nc++,ns++;
memset(pos,-1,sizeof pos);
btree(1,nc,1);
ans=-1;
for(i=0;i<nc;i++)
{
scanf("%d",&pc[i]);
pos[pc[i]]=i+1;//记录序列1每个数字的位置
}
for(i=0;i<ns;i++)
scanf("%d",&pcs[i]);
for(i=0;i<ns;i++)
{
if(pos[pcs[i]]==-1)
continue;
if(pos[pcs[i]]==1)
tp=1;
else
tp=qu(1,nc,1,pos[pcs[i]]-1,1)+1;//关键部分
update(1,nc,pos[pcs[i]],1,tp);
ans=max(tp,ans);
}
printf("Case %d: %d\n",cas++,ans);
}
return 0;
}