hdu 1511(dp)

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转载标记处：http://blog.csdn.net/my_acm_dream/article/details/41577645

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 90int 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;}

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