A - Prince and Princess Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu Submit Status Practice UVA 10635 Appoint description:
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 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
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
| 1 3 6 7 1 7 5 4 8 3 9 1 4 3 5 6 2 8 9 | Case 1: 4 |
求2個串的最長公共子序列
因為資料範圍是250*250 樸素算法會t
又因為每個串的每個數字出現的次數都為一
是以可以轉為求lis
如何轉換？
對于a數組{1,7,5,4,8,3,9}記錄每個數字出現的先後順序{1（1）,7（2）,5（3）,4（4）,8（5）,3（6）,9（7）}
然後在對b數組給出其資料在a數組出現的位置 b{1（a[1]=1),4(a[4]=4),3(a[3]=6),5(a[5]=3),6(a[6]=0),2(a[2]=0),8(a[8]=5),9(a[9]=7)}
然後對得到數組求lis {1,4,6,3,0,0,5,7} 可以看出是 1,3,5,7 長度是4
最長上升子序列（LIS），熟悉的n^2的動歸會逾時。LIS問題可以優化為nlogn的算法。
定義d[k]:長度為k的上升子序列的最末元素，若有多個長度為k的上升子序列，則記錄最小的那個最末元素。
注意d中元素是單調遞增的，下面要用到這個性質。
首先len = 1,d[1] = a[1],然後對a[i]:若a[i]>d[len]，那麼len++，d[len] = a[i];
否則，我們要從d[1]到d[len-1]中找到一個j，滿足d[j-1]<a[i]<d[j],則根據D的定義，我們需要更新長度為j的上升子序列的最末元素（使之為最小的）即 d[j] = a[i];
最終答案就是len
利用d的單調性，在查找j的時候可以二分查找，進而時間複雜度為nlogn。
ACcode：
#include <map>
#include <queue>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#define maxn 70000
using namespace std;
int a[maxn],b[maxn],dp[maxn];
int find_it(int x,int len){
int l,r,m;
l=0,r=len;
while(l<r){
m=(l+r)>>1;
if(dp[m]>=b[x])r=m;
else l=m+1;
}
return l;
}
int main(){
int t,n,la,lb,cnt=1,tmp;
scanf("%d",&t);
while(t--){
scanf("%d%d%d",&n,&la,&lb);
la++,lb++;
for(int i=1;i<=la;++i){
scanf("%d",&tmp);
a[tmp]=i;
}
for(int i=1;i<=lb;++i){
scanf("%d",&tmp);
b[i]=a[tmp];
}
tmp=1;
dp[1]=b[1];
for(int i=2;i<=lb;++i)
if(b[i]>dp[tmp])
dp[++tmp]=b[i];
else{
int pos=find_it(i,tmp);
dp[pos]=b[i];
}
printf("Case %d: %d\n",cnt++,tmp);
}
return 0;
}
/*
1
3 6 7
1 7 5 4 8 3 9
1 4 3 5 6 2 8 9
*/