回文串最少划分数

这个是正向分析字符串，觉得比较好理解.
这个问题以后还要补充……

via http://blog.youkuaiyun.com/u010885899/article/details/50364183

#pragma warning(disable:4996)  
#include <iostream>  
#include <algorithm>  
#include <cmath>  
#include <vector>  
#include <string>  
#include <cstring>  
using namespace std;

int len;
char s[5005];
char s1[5005];
int dp[5005];
int v[5005][5005];

void input()
{
    scanf("%s", s1);
    len = strlen(s1);
    strcpy(s + 1, s1);
}

void solve()
{
    int i, j;
    for (i = 0; i <= len; i++)
    {
        v[i][i] = 1;
    }
    for (i = 1; i <= len; i++)
    {
        dp[i] = i;
        for (j = i; j >= 1; j--)
        {
            if (i == j)
            {
                dp[i] = min(dp[i], dp[i - 1] + 1);
            }
            else if (i - j == 1 && s[i] == s[j])
            {
                v[j][i] = 1;
                dp[i] = min(dp[i], dp[j-1] + 1);
            }
            else if (v[j + 1][i - 1]&&s[i]==s[j])
            {
                v[j][i] = 1;
                dp[i] = min(dp[i], dp[j-1] + 1);
            }
        }
    }
    printf("%d", dp[len]);
}

int main()
{
    //freopen("i.txt", "r", stdin);
    //freopen("o.txt", "w", stdout);

    input();
    solve();

    //system("pause");
    return 0;
}