UVA - 11584 Partitioning by Palindromes（划分成回文串）（dp）

题意：输入一个由小写字母组成的字符串，你的任务是把它划分成尽量少的回文串，字符串长度不超过1000。

分析：

1、dp[i]为字符0~i划分成的最小回文串的个数。

2、dp[j] = Min(dp[j], dp[i - 1] + 1)，若i~j是回文串，则更新dp[j]。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 1000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
char s[MAXN];
int dp[MAXN];
bool judge(int l, int r){
    int len = r - l + 1;
    for(int i = 0; i < len / 2; ++i)
        if(s[l + i] != s[r - i]) return false;
    return true;
}
int main(){
    int T;
    scanf("%d", &T);
    while(T--){
        memset(dp, INT_INF, sizeof dp);
        scanf("%s", s + 1);
        int len = strlen(s + 1);
        dp[0] = 0;
        for(int i = 1; i <= len; ++i){
            for(int j = i; j <= len; ++j){
                if(judge(i, j)){
                    dp[j] = Min(dp[j], dp[i - 1] + 1);
                }
            }
        }
        printf("%d\n", dp[len]);
    }
    return 0;
}