uva 11584 Partitioning by Palindromes

原题：
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’).
Can you read upside-down?
• ‘aaadbccb’ can be partitioned as (‘aaa’, ‘d’, ‘bccb’).
Input
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.Universidad de Valladolid OJ: 11584 – Partitioning by Palindromes
Sample Input
3
racecar
fastcar
aaadbccb
Sample Output
1
7
3
题目大意：
给你一个字符串，问你这个字符串最少能拆分成多少个回文串的组合。

#include <bits/stdc++.h>using namespace std;//fstream in,out;bool mark[1001][1001];//用来记录i到j这段字符串是否是回文int dp[1001];string s;int main(){    ios::sync_with_stdio(false);    int n;    cin>>n;    while(n--)    {        cin>>s;        memset(dp,0,sizeof(dp));        memset(mark,false,sizeof(mark));        int len=s.size();        for(int i=0;i<len;i++)            mark[i][i]=true;        for(int i=0;i<len-1;i++)            if(s[i]==s[i+1])                mark[i+1][i]=mark[i][i+1]=true;//相邻的两个字符相同要先判断        for(int k=3;k<=len;k++)        {            for(int i=0;i<len-k+1;i++)            {                int j=k+i-1;                if(s[i]==s[j]&&mark[i+1][j-1])                    mark[j][i]=mark[i][j]=true;            }        }        dp[0]=1;        for(int i=1;i<len;i++)        {            if(mark[i][0])            {                dp[i]=1;                continue;            }            else                dp[i]=INT_MAX;            for(int j=1;j<i;j++)            {                if(mark[i][j])                    dp[i]=min(dp[j-1]+1,dp[i]);            }            dp[i]=min(dp[i],dp[i-1]+1);        }        cout<<dp[len-1]<<endl;    }    return 0;}

解答：
之前在leetcode上面做过一个最长回文串的题目，没想到在这用上了。用mark[i][j]表示i到j是否是回文串，如果s[i]==s[j]而且mark[i+1][j-1]为真，那么mark[i][]j]肯定为真，也就是i到j是一个回文串。
接下来设置dp[i]表示以字母s[i]为结尾的字符串最少能分解成多少个回文串，那么dp[i]=min(dp[j-1]+1,dp[i]) 当mark[i][j]为真的情况下。
转移方程的意思是如果mark[i][j]为真，那么就把字符串j前面字符能组成最少回文串数加上回文串i到j即可。