lightoj-1044 - Palindrome Partitioning(区间dp)

1044 - Palindrome Partitioning
PDF (English) Statistics Forum
Time Limit: 1 second(s) Memory Limit: 32 MB
A palindrome partition is the partitioning of a string such that each separate substring is a palindrome.

For example, the string "ABACABA" could be 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 are 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 you have to print the case number and the desired result.

Sample Input
Output for Sample Input
3
AAAA
ABCDEFGH
QWERTYTREWQWERT
Case 1: 1
Case 2: 8
Case 3: 5

 

解题:预处理好回文字符串，然后dfs。

但dp[l][r] = min(dp[l][r],dfs(l,i)+dfs(i+1,r)); 这种的dfs会爆

仔细分析下会发现 (回文字符串+一串字母) + 一串字母 跟 回文字符串+(一串字母+一串字母) 的情况是一样， 所以可能省略掉前面dfs(l,i)的情况

直接  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 namespace std;

const int inf = 1e9;
int T,len;
char str[1100];
int dp[1100][1100];
bool judge[1100][1100];

void init(){
    memset(judge,false,sizeof(judge));
    memset(dp,-1,sizeof(dp));    
    
    for(int i=0;i<len;i++){
        for(int j=i;j<len;j++){
            int flag = 0;
            for(int z=i;z<=(i+j)/2;z++){
                if(str[z]!=str[j-(z-i)]){
                    flag = 1;break;
                }
            }
            if(flag == 0) judge[i][j] = true;
        }
    }    
}

int dfs(int l,int r){
    
    if(judge[l][r]){
        dp[l][r] = 1;
        return 1;
    }
    if(dp[l][r]!=-1) return dp[l][r];
    
    dp[l][r] = inf;
    for(int i=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));        
    }
    return dp[l][r];
}

int main(){
    
    scanf("%d",&T);
    for(int t=1;t<=T;t++){
        
        scanf("%s",str);
        len = strlen(str);
        init();
        printf("Case %d: %d\n",t,dfs(0,len-1));
    }
    
    return 0;
}