poj2004——Mix and Build

思路：用长度作为关键字排序，对于str[i].ch，寻找i之前能被str[i].ch继承的最大的str[i-x].ch！

注意点：考虑aa,aaa,aaaa的情况

#include<iostream>#include<cstdio>#include<string>#include<algorithm>using namespace std;class node{public:char ch[25];int vis[29];int len;int par;int t;node(){par=-1;t=0;memset(vis,0,sizeof(vis));}};node str[100005];char ch[25],pri[100005][25];int cnt;bool cmp(node &a,node &b){if(a.len <b.len ) return true;return false;}bool isexten(int a,int b){int i,len=str[b].len ;for(i=0;i<len;i++){int k=str[b].ch [i]-'a';if(str[a].vis [k]<str[b].vis [k]) return false;}return true;}int par[10005];void build(){int i,j;int maxlen=0,index=0;int star,end;end=cnt-1;star=1;while(star<=end){i=star-1;while(i>=0 ){int dis=str[star].len -str[i].len ;if(dis==0) {}else {if(dis>1) break;if(isexten(star,i)){if(str[star].t <str[i].t +1){str[star].t=str[i].t +1;str[star].par =i;}}}i--;}if(str[star].t >maxlen){maxlen=str[star].t ;index=star;}star++;}int k=index ,pr=0;while(k!=-1){strcpy(pri[pr++],str[k].ch );k=str[k].par ;}for(i=pr-1;i>=0;i--)printf("%s/n",pri[i]);}int main(){cnt=0;while(scanf("%s",ch)!=EOF){strcpy(str[cnt].ch ,ch);str[cnt].len =strlen(ch);for(int i=0;i<str[cnt].len ;i++){str[cnt].vis [str[cnt].ch [i]-'a']++;}cnt++;}sort(str,str+cnt,cmp);build();return 0;}