POJ 题目1239 ||ZOJ 题目 1499 Increasing Sequences（正反两次DP）

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Increasing Sequences

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 3025 Accepted: 1147

Description

Given a string of digits, insert commas to create a sequence of strictly increasing numbers so as to minimize the magnitude of the last number. For this problem, leading zeros are allowed in front of a number.

Input

Input will consist of multiple test cases. Each case will consist of one line, containing a string of digits of maximum length 80. A line consisting of a single 0 terminates input.

Output

For each instance, output the comma separated strictly increasing sequence, with no spaces between commas or numbers. If there are several such sequences, pick the one which has the largest first value;if there's a tie, the largest second number, etc.

Sample Input

34563546352600011000001010

Sample Output

3,4,5,635,463,5,260001100,000101

Source

East Central North America 2002

思路：http://blog.csdn.net/vecri/article/details/4758786

题目大意： 给定一个字符串， 如3456, 将其分割成多个整数，使该整数序列递增，且尽可能使最大的数也就是序列最后一个数最小，在这个前提下使序列前面的数最大分析：两次DP, 一次前向DP，一次后向DP第一次DP来确定最后一个数字，因为这个数字是递增序列的最后一个数，且这个数必须最小,代码中dp[i]=j是指由str[1...i]序列生成的递增序列的最后一个数是str[j...i]第二次DP是在确定最后一个数字的基础上，往前规划，使得前面的数尽可能的大，代码中dp2[i]=j是指在确定最后一个数的情况下，str[i...end]序列中最大的开头数为str[i...j]注意最后一个数前面可以加零。

ac代码

#include<stdio.h>#include<string.h>char str[110],s[110];int dp[110],dp2[110];int cmp(int b1,int e1,int b2,int e2){while(str[b1]=='0')b1++;while(str[b2]=='0')b2++;int len1=e1-b1+1;int len2=e2-b2+1;if(len1<len2)return -1;elseif(len1>len2)return 1;elsereturn strncmp(str+b1,str+b2,len1);}int main(){while(scanf("%s",str)!=EOF){if(strcmp(str,"0")==0)break;memset(dp,0,sizeof(dp));memset(dp2,0,sizeof(dp2));int len=strlen(str);int i,j;for(i=1;i<len;i++){for(j=i-1;j>=0;j--){if(cmp(dp[j],j,j+1,i)<0){dp[i]=j+1;break;}}}i=dp[len-1];dp2[i]=len-1;while(str[i-1]=='0'){dp2[i-1]=len-1;--i;}for(i=dp[len-1]-1;i>=0;i--){for(j=i;j<dp[len-1];j++){if(cmp(i,j,j+1,dp2[j+1])<0&& dp2[i]<j){dp2[i]=j;}}}i=0;while(i<len){strncpy(s,str+i,dp2[i]-i+1);s[dp2[i]-i+1]='\0';if(i!=0)printf(",");printf("%s",s);i=dp2[i]+1;}printf("\n");}}

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