hdu 1511(dp)

解题思路：两次dp确实很巧妙，我只想到了一次dp算出最后那个数最小，其实题目要求还要保证第一个数尽可能大，一开始也没有注意到。。

AC:

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<math.h>
#include<stdlib.h>
#include<time.h>
using namespace std;
#define oo 0x3f3f3f3f
#define maxn 90
int dp_min[maxn];
int dp_max[maxn];

int smaller(char *s1,char *e1,char *s2,char *e2)
{
    while(*s1 == '0' && s1 != e1) ++s1;
    while(*s2 == '0' && s2 != e2) ++s2;
    int len1 = e1-s1;
    int len2 = e2-s2;
    if(len1 != len2)
        return len1-len2;
    while(s1 != e1)
    {
        if(*s1 != *s2)
            return *s1-*s2;
        ++s1;
        ++s2;
    }
    return 1;
}

int main()
{
    char str[maxn];
    while(scanf("%s",str+1) && strcmp(str+1,"0"))
    {
        int n=strlen(str+1);
        for(int i = 1;i <= n;i++)
            dp_min[i] = dp_max[i] = 1;	//注意这个初始化，开始时每个数结尾的范围都是1-dp[i];
        for(int i = 2;i <= n;i++)
        {
            for(int j = i-1;j >= 1; j--)
            {
                if(smaller(str+dp_min[j],str+j+1,str+j+1,str+i+1) < 0)
                {
                    dp_min[i]=j+1;
                    break;
                }
            }
        }

        int m = dp_min[n];
        dp_max[m] = n;
        int i,j;
        for(i = m-1; i >= 1 && str[i] == '0'; i--)
            dp_max[i] = n;
        for(; i >= 1; i--)
        {
            for(j = m-1; j >= i; j--)
            {
                if(smaller(str+i,str+j+1,str+j+1,str+dp_max[j+1]+1) < 0)
                {
                    dp_max[i] = j;
                    break;
                }
            }
        }

        int f = 0;
        for(i = 1,j = dp_max[i];i <= n;j = dp_max[j+1])
        {
            if(!f) f = 1;
            else printf(",");
            while(i <= j && i <= n)
                printf("%c",str[i++]);
        }
        puts("");
    }
	return 0;
}