HDU 4745 interval DP

Test instructions: The longest palindrome sequence of a ring, is a sequence not a string

Link: Point Me

The starting point can be arbitrary,

So only the longest palindrome sequence of each interval is required to take Max (Dp[1][i]+dp[i+1][n]), the final answer can be

Originally wanted to expand twice times, but later the biggest not too good to think

1. Multiply the ring into a chain, find the window is the eldest of N sequence, but this is not the final solution, you can try to see the sample 2, is only 4, because it can be selected in the palindrome outside also select as a starting point to jump, so the outside to determine whether the palindrome outside can also be the starting point of the stone, that can find the window (n-1) of length +1. So the solution is to find the length of the window n or the maximum value of the window (n-1) length +1.

2. Do not multiply, directly as a chain to seek DP, and then cut into two halves, to find the length of the palindrome on both sides, the largest and is the solution. There is no need to consider the starting point, because both sides of the palindrome can do the starting point.

Sample Input

1141 1 2 162 1 1 2 1 30
Sample Output

145



2015-05-06 Code

1#include <cstdio>2#include <iostream>3#include <algorithm>4#include <cstring>5#include <cmath>6#include <queue>7#include <map>8 using namespacestd;9 #defineMOD 1000000007Ten Const intinf=0x3f3f3f3f; One Const Doubleeps=1e-5; AtypedefLong Longll; - #defineCL (a) memset (A,0,sizeof (a)) - #defineTS printf ("*****\n"); the Const intmaxn=1005; - intN,m,tt; - intA[MAXN]; - intDP[MAXN][MAXN]; + intMain () - { +     inti,j,k; A #ifndef Online_judge atFreopen ("1.in","R", stdin); -     #endif -      while(SCANF ("%d", &n)! =EOF) -     { -         if(n==0) Break; - CL (DP); in          for(i=0; i<n;i++) -         { toscanf"%d",&a[i]); +dp[i][i]=1; -         } the          for(intlen=1; len<n;len++) *         { $              for(i=0; i+len<n;i++)Panax Notoginseng             { -j=len+i; the                 if(A[i]==a[j]) dp[i][j]=dp[i+1][j-1]+2; +                 ElseDp[i][j]=max (dp[i][j-1],dp[i+1][j]); A             } the         } +         intcn1=0; -          for(intI=0; i<n;i++) $Ans=max (ans,dp[0][i]+dp[i+1][n-1]); $printf"%d\n", ans); -     } -}

View Code

1#include <cstdio>2#include <algorithm>3#include <cstring>4 using namespacestd;5 intdp[1001][1001];6 inta[1001];7 intN;8 intMain ()9 {Ten      while(SCANF ("%d", &n) &&N) One     { A          for(intI=1; i<=n;i++) -scanf"%d",&a[i]); -Memset (DP,0,sizeof(DP)); the          for(intI=1; i<=n;i++) -dp[i][i]=1; -          for(intlen=1; len<n;len++) { -              for(intI=1; i+len<=n;i++) { +                 intj=i+Len; -                 if(a[i]==A[j]) +dp[i][j]=dp[i+1][j-1]+2; A                 Else atDp[i][j]=max (dp[i][j-1],dp[i+1][j]); -             } -         } -         intcn1=0; -          for(intI=1; i<=n;i++) -Ans=max (ans,dp[1][i]+dp[i+1][n]); inprintf"%d\n", ans); -     } to     return 0; +}

HDU 4745 interval DP