ural 1002 phone numbers

这是一道动归题, 算法复杂度为 O(n*len).

#include<iostream>using namespace std;const int MAX=10000;char str[105];int n;char s[50002][55];int f[105][2];int l,len[50002];int T[26]={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};///////////a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,zint cmp(int j,int i){ int t; for(t=0;t<len[j];t++) { if(str[i+1-len[j]+t]-'0'!=T[s[j][t]-'a']) return 0; } return 1;}void print(int i){ if(i<=0) return; print(i-len[f[i][1]]); printf("%s",s[f[i][1]]); if(i!=l-1) printf(" "); else printf("/n"); return;}int main(){ int i,j,k; while(scanf("%s",str)&&str[0]!='-') { l=strlen(str); scanf("%d",&n); for(i=1;i<=n;i++) { scanf("%s",s[i]); len[i]=strlen(s[i]); } for(i=0;i<l;i++) { f[i][0]=MAX; for(j=1;j<=n;j++) { if(len[j]>i+1) continue; if(len[j]==i+1) { if(cmp(j,i)) { f[i][0]=1; f[i][1]=j; } } else if(cmp(j,i)) { if(f[i-len[j]][0]+1<f[i][0]) {f[i][0]=f[i-len[j]][0]+1;f[i][1]=j; }} }}if(f[l-1][0]==MAX) printf("No solution./n");else{ print(l-1);}memset(s,0,sizeof(s));memset(f,0,sizeof(f));memset(str,0,sizeof(str));memset(len,0,sizeof(len)); } return 0;}