Ural 1002 Phone number

dp[i]表示数字串第i位之前最少有多少单词,以数字串长度为阶段,每个阶段共有n个状态,类似01背包..不同的是,单词可以放不止一次...

转移方程dp[i]=min(dp[i],dp[i-word[j].length()]+1);关键是记录,dp[i]每更新一次状态,记录下j,最后一个j肯定用单词最短的那个单词,接着向前遍历.....

#include<iostream>#include<cstring>#include<string>using namespace std;#define MAX 50000string word[MAX+2];int res[MAX+2][52];int dp[102]; string num;int a[102];int ans[102];int  w[102];const int f[27]={2,2,2,3,3,3,4,4,1,1,5,5,6,6,0,7,0,7,7,8,8,8,9,9,9,0};bool fun(int pos,int st){    int j=0;    for(int i=pos-word[st].length();i<pos;i++)        if(a[i]!=res[st][j++]) return false;    return true;}int main(){      int n,nlen,i,j,t;   while(cin>>num&&num[0]!='-')   {    nlen=num.length();    for(i=0;i<nlen;i++) a[i]=num[i]-'0';    cin>>n;     for(i=0;i<n;i++)     {     cin>>word[i];     t=word[i].length();     for(j=0;j<t;j++)        res[i][j]=f[word[i][j]-'a'];     }    for(i=0;i<=nlen;i++) {dp[i]=1<<30;w[i]=0;}    dp[0]=0;    for(i=1;i<=nlen;i++)    {     for(j=0;j<n;j++)         if(i>=word[j].length()&&fun(i,j)&&dp[i-word[j].length()]+1<dp[i])         {dp[i]=dp[i-word[j].length()]+1;          w[i]=j;        }    }    if(dp[nlen]>2000)    printf("No solution.\n");    else    {     i=nlen;int s=0;     while(i>0)     {       ans[s++]=w[i];       i-=word[w[i]].length();     }    for(i=s-1;i>0;i--)        cout<<word[ans[i]]<<" ";    cout<<word[ans[0]]<<endl;    }   }}