Partitioning by Palindromes

题意：

给定一个字符串，求能分成最小几个回文串

分析：简单dp dp[i]前i个字符能分成的最小数量 dp[i]=min(dp[i],dp[j-1]+1) (j-i 是回文串）

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <vector>
#include <string>
#include <cctype>
#include <complex>
#include <cassert>
#include <utility>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
typedef pair<int,int> PII;
typedef long long ll;
#define lson l,m,rt<<1
#define pi acos(-1.0)
#define rson m+1,r,rt<<11
#define All 1,N,1
#define read freopen("in.txt", "r", stdin)
const ll  INFll = 0x3f3f3f3f3f3f3f3fLL;
const int INF= 0x7ffffff;
const int mod =  ;
int dp[];
char s[];
int judge(int x,int y){
    while(x<y){
        if(s[x]!=s[y])return ;
        x++;y--;
    }
    return ;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%s",(s+));
        int len=strlen(s+);
        for(int i=;i<=len;++i)
            dp[i]=INF;
            dp[]=;
        for(int i=;i<=len;++i)
            for(int j=;j<=i;++j)
            if(judge(j,i)){
                dp[i]=min(dp[i],dp[j-]+);
            }
        printf("%d\n",dp[len]);
    }
return ;
}