Lightoj-1044-palindrome partitioning (interval DP)

1044-palindrome Partitioning
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Time Limit:1 second (s) Memory limit:32 MB
A palindrome partition is the partitioning of a string such, each separate substring is a palindrome.

For example, the string "Abacaba" could is partitioned in several different ways, such as {"A", "B", "a", "C", "a", "B", "a"}, { "A", "Bacab", "a"}, {"ABA", "C", "ABA"}, or {"Abacaba"}, among others.

You is given a string s. Return the minimum possible number of substrings in a palindrome partition of S.

Input
Input starts with an integer T (≤40), denoting the number of test cases.

Each case begins with a non-empty string s of uppercase letters with length no more than 1000.

Output
For each case of input, the case number and the desired result has been to print.

Sample Input
Output for Sample Input
3
Aaaa
Abcdefgh
Qwertytrewqwert
Case 1:1
Case 2:8
Case 3:5

Problem solving: Preprocess a palindrome string and then Dfs.

But dp[l][r] = min (Dp[l][r],dfs (l,i) +dfs (i+1,r)); This kind of DFS will explode

Careful analysis will find (Palindrome string + a string of letters) + a string of letters and palindrome string + (a string of letters + a string of letters) is the same situation, so may omit the previous DFS (l,i) situation

Direct if (Judge[l][i]) dp[l][r] = min (Dp[l][r],1+dfs (i+1,r));

#include <iostream>#include<cstring>#include<cstdio>#include<algorithm>using namespacestd;Const intINF =1e9;intT,len;Charstr[1100];intdp[1100][1100];BOOLjudge[1100][1100];voidinit () {memset (judge,false,sizeof(judge)); Memset (DP,-1,sizeof(DP));  for(intI=0; i<len;i++){         for(intj=i;j<len;j++){            intFlag =0;  for(intz=i;z<= (I+J)/2; z++){                if(Z-str[z]!=str[j-i)]) {Flag=1; Break; }            }            if(Flag = =0) Judge[i][j] =true; }    }    }intDfsintLintR) {        if(Judge[l][r]) {Dp[l][r]=1; return 1; }    if(dp[l][r]!=-1)returnDp[l][r]; DP[L][R]=inf;  for(inti=l;i<r;i++){                if(Judge[l][i]) Dp[l][r]= Min (Dp[l][r],1+dfs (i+1, R)); //Dp[l][r] = min (Dp[l][r],dfs (l,i) +dfs (i+1,r));     }    returndp[l][r];}intMain () {scanf ("%d",&T);  for(intt=1; t<=t;t++) {scanf ("%s", str); Len=strlen (str);        Init (); printf ("Case %d:%d\n", T,dfs (0, len-1)); }        return 0;}

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Lightoj-1044-palindrome partitioning (interval DP)