UVA 11584 Partitioning by Palindromes

题意：把字符串划分成尽量少的回文串。

dp[i] = max{dp[j-1] + 1 | str[j....i]为回文串}。

/*
Author:Zhaofa Fang
Lang:C++
*/
#include <cstdio>
#include <cstdlib>
#include <sstream>
#include <iostream>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <string>
#include <utility>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
using namespace std;

typedef long long ll;
#define DEBUG(x) cout<< #x << ':' << x << endl
#define REP(i,n) for(int i=0;i < (n);i++)
#define FOR(i,s,t) for(int i = (s);i <= (t);i++)
#define PII pair<int,int>
#define PB push_back
#define MP make_pair
#define FI first
#define SE second
#define lowbit(x) (x&(-x))
#define INF (1<<30)

char str[1010];
int dp[1010];

bool check(char s[],int l,int r)
{
    for(int i=l,j=r;i<=r;i++,j--)
        if(s[i] != s[j])return false;
    return true;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("in","r",stdin);
#endif
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s",str+1);
        int len = strlen(str+1);
        REP(i,len+1)dp[i] = INF;
        dp[0] = 0;
        FOR(i,1,len)
        {
            FOR(j,1,i)
            {
                if(check(str,j,i))
                {
                    dp[i] = min(dp[i],dp[j-1] + 1);
                }
            }
        }
        printf("%d\n",dp[len]);
    }
    return 0;
}