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 is numbered 1, 2, 3 ... n*n, as shown below:
Prince stands in square 1 , make P jumps and finally reach square& nbsp N*n . He enters a square at the most once. So if we use x p to denote the P -thsquare he enters, then x 1 , x2 , ... xp+1 is all different. Note that x 1 = 1 and x P + 1 = n*n . Princess does the similar thing-stands in square 1 , make Q jumps and F Inally reach square n*n . We use y 1 , y2 , ... yq+1 to denote the sequence, and all q+1 numbers is 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--and 5--and 4---8---3 (Black arrows), whil E The princess moves along this sequence:1 --4--and 3--and 5---6--and 2--and 8 (white Arrow).
The King--their father, has just come. "Why move separately?" You're brother and sister! "said the King," Ignore some jumps and make sure this you ' re always together. "
For example, if the Prince ignores him 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 your tell him?
Input
The first line of the input contains a single integer t (1 <= t <=), the number of test cases followed. For each case, the first line contains three integers n, p, q (2 <= n <=, 1 <= p, q < n*n). The second line contains p+1 different integers in the range [1..n*n], the sequence of the Prince. The third line contains q+1 different integers in the range [1..n*n], the sequence of the princess.
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 |
Lcs-->lis
#include <iostream>#include<cstdio>#include<cstring>using namespacestd;#defineN 62555intLen;intn,p,q;intA[n];intB[n];intDp[n];voidConvert ()//Convert a sequence to 1--p,b sequence mapping.{ inti,hash[n]={0}; for(i=1; i<=p;i++) {Hash[a[i]]=i; } for(i=1; i<=q;i++) {B[i]=Hash[b[i]]; }}intUp_bound (intk) { intL=1, r=len+1; while(l<r) {intM= (l+r) >>1; if(dp[m]<=k) l=m+1; ElseR=m; } returnl;}voidsolve () {Len=1; dp[1]=b[1]; for(intI=2; i<=q;i++) { if(B[i]>dp[len]) dp[++len]=B[i]; Else { intpos=Up_bound (B[i]); Dp[pos]=B[i]; }} printf ("%d\n", Len);}intMain () {intT,i,icase=1; scanf ("%d",&T); while(t--) {scanf ("%d%d%d",&n,&p,&q); P++; Q++; for(i=1; i<=p;i++) {scanf ("%d",&A[i]); } for(i=1; i<=q;i++) {scanf ("%d",&B[i]); } convert (); printf ("Case %d:", icase++); Solve (); } return 0;}
Algorithmic race and Getting started classic---P66 [UVA 10635] Prince and Princess