UVA - 11584 Partitioning by Palindromes[序列DP]

Posted 2020-08-15 Candy?

UVA - 11584 Partitioning by Palindromes

We say a sequence of char- acters is a palindrome if it is the same written forwards and backwards. For example, ‘racecar’ is a palindrome, but ‘fastcar’ is not.

A partition of a sequence of characters is a list of one or more disjoint non-empty groups of consecutive characters whose concatenation yields the initial sequence. For example, (‘race’, ‘car’) is a partition of ‘racecar’ into two groups.

Given a sequence of charac- ters, we can always create a par- tition of these characters such that each group in the partition is a palindrome! Given this ob- servation it is natural to ask: what is the minimum number of groups needed for a given string such that every group is a palin- drome?

For example:

• ‘racecar’ is already a palindrome, therefore it can be partitioned into one group.

• ‘fastcar’ does not con- tain any non-trivial palin- dromes, so it must be par- titioned as (‘f’, ‘a’, ‘s’, ‘t’, ‘c’, ‘a’, ‘r’).

• ‘aaadbccb’ can be parti- tioned as (‘aaa’, ‘d’, ‘bccb’).

  Input

Can you read upside-down?

Input begins with the number n of test cases. Each test case consists of a single line of between 1 and 1000 lowercase letters, with no whitespace within.

Output

For each test case, output a line containing the minimum number of groups required to partition the input into groups of palindromes.

Sample Input

3
racecar
fastcar
aaadbccb

Sample Output

1 7 3 

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最少回文划分

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f[i]表示到i的最少次数，随便一转移

判断回文：p[i][j]=p[i+1][j-1] if s[i]==s[j]

//
//  main.cpp
//  uva11584
//
//  Created by Candy on 10/18/16.
//  Copyright © 2016 Candy. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
const int N=1005,INF=1e9;
inline int read(){
    char c=getchar();int x=0,f=1;
    while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1;c=getchar();}
    while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘;c=getchar();}
    return x*f;
}
int T,n;
char s[N];
int f[N],p[N][N];
bool pal(int l,int r){
    if(l>=r) return 1;
    if(p[l][r]!=-1) return p[l][r];
    if(s[l]==s[r]) p[l][r]=pal(l+1,r-1);
    else p[l][r]=0;
    return p[l][r];
}
void dp(){
    memset(p,-1,sizeof(p));
    f[0]=0;f[1]=1;
    for(int i=2;i<=n;i++){
        f[i]=INF;
        for(int j=0;j<i;j++)
            if(pal(j+1,i)) f[i]=min(f[i],f[j]+1);
    }
}
int main(int argc, const char * argv[]) {
    T=read();
    while(T--){
        scanf("%s",s+1);
        n=strlen(s+1);
        dp();
        printf("%d\n",f[n]);
    }

    return 0;
}