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 x_p to denote the p-thsquare he enters, then x₁, x₂, ... x_p+1 are all different. Note that x₁ = 1 and x_p+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 , ... y_q+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

LCS-->LIS

#include <iostream>#include <cstdio>#include <cstring>using namespace std;#define N 62555int len;int n,p,q;int a[N];int b[N];int dp[N];void convert()             //将a序列转化为1--p，b序列映射过来{    int i,hash[N]={0};    for(i=1;i<=p;i++)    {        hash[a[i]]=i;    }    for(i=1;i<=q;i++)    {        b[i]=hash[b[i]];    }}int up_bound(int k){    int l=1,r=len+1;    while(l<r)    {        int m=(l+r)>>1;        if(dp[m]<=k) l=m+1;        else r=m;    }    return l;}void solve(){    len=1;    dp[1]=b[1];    for(int i=2;i<=q;i++)    {        if(b[i]>dp[len]) dp[++len]=b[i];        else        {            int pos=up_bound(b[i]);            dp[pos]=b[i];        }    }    printf("%d\n",len);}int main(){    int T,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;}

算法竞赛与入门经典---P66 [UVA 10635] Prince and Princess