UVA - 10635 Prince and Princess

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本文讨论了如何利用最长上升子序列（LIS）算法解决特定问题，并提供了具体的代码实现。通过实例分析，展示了LIS算法在解决路径规划问题中的应用，包括输入输出规范、案例输入输出示例及问题解析。

Prince and Princess

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

Submit Status

Description

Problem E
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-th square 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

The first line of the input contains a single integer t(1 <= t <= 10), the number of test cases followed. For each case, the first line contains three integers n, p, q(2 <= n <= 250, 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

3 6 7

1 7 5 4 8 3 9

1 4 3 5 6 2 8 9

Case 1: 4

	Problemsetter: Man Rujia Liu Man, Member of Elite Problemsetters' Panel

	Pictures drawn by Shahriar Manzoor, Member of Elite Problemsetters' Panel

分析：

刘汝佳大白书把它归到dp，其实最重要的是求解LIS（最长上升子序列）。用了O(nlogn)的算法。就是用了二分，调用了一个库函数。

LIS还是值得好好看看的，特别是nlogn的算法。

ac代码：

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=250*250+10;
const int inf=0x7fffffff;
int s[maxn],g[maxn],d[maxn];//d[maxn]可以省
int num[maxn];

int main()
{
int t;
scanf("%d",&t);
for(int kase=1;kase<=t;kase++)
{
int n,p,q,x;
scanf("%d%d%d",&n,&p,&q);
memset(num,0,sizeof(num));
for(int i=1;i<=p+1;i++)
{
scanf("%d",&x);
num[x]=i;//关联x，赋予编号i
}
int m=0;
for(int i=0;i<q+1;i++)
{
scanf("%d",&x);
if(num[x]) s[m++]=num[x];//p中没出现的元素全部删除
}
for(int i=1;i<=m;i++)
g[i]=inf;
int ans=0;
for(int i=0;i<m;i++)
{
int k=lower_bound(g+1,g+1+m,s[i])-g;
g[k]=s[i];
ans=max(ans,k);
}
printf("Case %d: %d\n",kase,ans);
}
return 0;
}