UVA 11584

Problem H: Partitioning by Palindromes

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').
• 'aaadbccb' can be partitioned as ('aaa', 'd', 'bccb').

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.

For each test case, output a line containing the minimum number of groups required to partition the input into groups of palindromes.

Sample Input

3
racecar
fastcar
aaadbccb

Sample Output

1
7
3

---

Kevin Waugh

这道题倒是没什么可说的，简单的DP入门题 ， 但我却在题以外的地方上WA了好久

谁能告诉我ios::sync_with_stdio(false);再用cin ,cout 会判错！！！求科普！

悬赏解答该问题！

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <map>
#include <string>
#include <stack>
#include <cctype>
#include <vector>
#include <queue>
#include <set>

using namespace std;
const int MAXN = 1010 + 50;
const int maxw = 100 + 20;
const long long LLMAX = 0x7fffffffffffffffLL;
const long long LLMIN = 0x8000000000000000LL;
const int INF = 0x7fffffff;
const int IMIN = 0x80000000;
#define eps 1e10-8
#define mod 1000000007
typedef long long LL;
const double PI = acos(-1.0);
typedef double D;

//#define Online_Judge
#define outstars cout << "***********************" << endl;
#define clr(a,b) memset(a,b,sizeof(a))

#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)
#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)
#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)
#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)
char inp[MAXN];
int state[MAXN];///记录从头到i的最少回文串数
bool isPalindrome(int s , int t)
{
    while(s < t)
    {
        if(inp[s] != inp[t])return false;
        s++;
        t--;
    }
    return true;
}
int main()
{
    //ios::sync_with_stdio(false);
    #ifdef Online_Judge
        freopen("in.txt","r",stdin);
        freopen("out.txt","w",stdout);
    #endif // Online_Judge
    int T;
    cin >> T;
    while(T--)
    {
        scanf("%s",inp);
        int len = strlen(inp);
        FOR(i , 0 , len)state[i] = INF;
        state[0] = 1;
        FOR(i , 1 , len)FORR(j ,0 , i)
        {
            if(isPalindrome(j , i))
            {
                if(j == 0)state[i] = min(state[i] , 1);
                else state[i] = min(state[j - 1] + 1 , state[i]);
            }
        }
        printf("%d
" , state[len - 1]);
    }
    return 0;
}