poj 1239 Increasing Sequences dp lis

Increasing Sequences

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 3081 Accepted: 1174

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

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求把一个数划分若干个数然后要满足构成数是递增的 在保证最后一个数最小的情况下第一个数最大

ACcode

#include <iostream>#include <cstdio>#include <cstring>#define maxn 100using namespace std;int dp[maxn];char str[maxn];bool big(int i,int j,int m,int n){    while(str[i]=='0'&&i<=j)i++;    while(str[m]=='0'&&m<=n)m++;    if(i>j)return false;    if(m>n)return true;    int a=j-i+1,b=n-m+1;    if(a>b)return true;    if(a<b)return false;    for(int t=i,p=m;t<=j&&p<=n;++p,++t)        if(str[t]>str[p])return true;        else if(str[t]<str[p])return false;    return false;}int main(){    while(scanf("%s",str+1)!=EOF){        int len=strlen(str+1);        if(len==1&&str[1]=='0')break;        for(int i=1;i<=len;++i)dp[i]=i;        for(int i=2;i<=len;++i)            for(int j=i-1;j>0;--j)                if(big(j+1,i,j-dp[j]+1,j)){                    dp[i]=i-j;                    break;                }        int pos=len-dp[len]+1;        dp[pos]=dp[len];        for(int i=pos-1;i>0;--i){            if(str[i]=='0'){                dp[i]=dp[i+1]+1;                continue;            }            for(int j=pos;j>i;--j)                if(big(j,j+dp[j]-1,i,j-1)){                    dp[i]=j-i;                    break;                }        }      //  for(int i=1;i<=len;++i)cout<<dp[i]<<' ';        //cout<<'\12';        for(int i=1;i<=dp[1];++i)putchar(str[i]);        pos=dp[1]+1;        while(pos<=len){            putchar(',');            for(int i=pos;i<dp[pos]+pos;++i)putchar(str[i]);            pos=pos+dp[pos];        }        putchar('\n');    }    return 0;}