Uva 11584 - Partitioning by Palindromes dp

Problem H: Partitioning by Palindromes

We say a sequence of characters 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 characters, we can always create a partition of these characters such that each group in the partition is a palindrome! Given this observation it is natural to ask: what is the minimum number of groups needed for a given string such that every group is a palindrome?

For example:

• 'racecar' is already a palindrome, therefore it can be partitioned into one group.
• 'fastcar' does not contain any non-trivial palindromes, so it must be partitioned as ('f', 'a', 's', 't', 'c', 'a', 'r').
• 'aaadbccb' can be partitioned as ('aaa', 'd', 'bccb').

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.

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

Sample Input

3racecarfastcaraaadbccb

Sample Output

173

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Kevin Waugh

初学dp先从简单题开始~~~~~

题意：给出长度不超过1000的字符串，把它分割成若干个回文字串，求能分成的最少字串数。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <map>#include <stack>#include <vector>#include <set>#include <queue>#pragma comment (linker,"/STACK:102400000,102400000")#define maxn 1005#define MAXN 2005#define mod 1000000009#define INF 0x3f3f3f3f#define pi acos(-1.0)#define eps 1e-6#define lson rt<<1,l,mid#define rson rt<<1|1,mid+1,rtypedef long long ll;using namespace std;int dp[maxn];char str[maxn];bool ISok(int s,int e){    for (int i=s;i<=s+(e-s+1)/2;i++)        if (str[i]!=str[e+s-i])            return false;    return true;}int main(){    int n;    scanf("%d",&n);    while (n--)    {        memset(dp,0,sizeof(dp));        scanf("%s",str+1);        int len=strlen(str+1);        for (int i=1;i<=len;i++)        {            dp[i]=i;            for (int j=1;j<=i;j++)            {                if (ISok(j,i))                    dp[i]=min(dp[i],dp[j-1]+1);            }        }        printf("%d\n",dp[len]);    }    return 0;}/*3racecarfastcaraaadbccb*/