UVa11584Partitioning by Palindromes(字符串区间dp)

题意：给定一个字符串s, 问说最少可以划分成几个回文串。

思路：dp[i]表示从1到第i个字符最少可以划分为几个回文，状态转移方程

dp[i] = min(dp[i], dp[j-1]+1),  如果满足 s[j] 到 s[i] 为回文字符串。

用 judge 函数判断从j到i是否可以形成回文串。

Sample Input

3

racecar

fastcar

aaadbccb

Sample Output

1

7

3

<span style="font-size:18px;">#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <string>
#include <algorithm>
#include <queue>
#include <stack>
using namespace std;

const double PI = acos(-1.0);
const double e = 2.718281828459;
const double eps = 1e-8;
const int MAXN = 1010;
char s[MAXN];
int dp[MAXN];

int judge(int l, int r)
{
    while(l <= r)
    {
        if(s[l] != s[r])
            return 0;
        l++;
        r--;
    }
    return 1;
}

int main()
{
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int Case;
    cin>>Case;
    while(Case--)
    {
        scanf("%s", s+1);
        int len = strlen(s+1);
        dp[0] = 0;
        for(int i = 1; i <= len; i++)
            dp[i] = 1010;
        for(int i = 1; i <= len; i++)
        {
            for(int j = 1; j <= i; j++)
            {
                if(judge(j, i))
                    dp[i] = min(dp[i], dp[j-1]+1);
            }
        }
        cout<<dp[len]<<endl;
    }
    return 0;
}</span>