POJ 2564 Edit Step Ladders

Edit Step Ladders

Time Limit: 3000MS Memory Limit: 131072KTotal Submissions: 2513 Accepted: 940

Description

An edit step is a transformation from one word x to another word y such that x and y are words in the dictionary, and x can be transformed to y by adding, deleting, or changing one letter. So the transformation from dig to dog or from dog to do are both edit steps. An edit step ladder is a lexicographically ordered sequence of words w1, w2, ... wn such that the transformation from wi to wi+1 is an edit step for all i from 1 to n-1.

Input

For a given dictionary, you are to compute the length of the longest edit step ladder. The input to your program consists of the dictionary - a set of lower case words in lexicographic order - one per line. No word exceeds 16 letters and there are no more than 25000 words in the dictionary.

Output

The output consists of a single integer, the number of words in the longest edit step ladder. 

Sample Input

catdigdogfigfinfinefoglogwine

分析：DP，最长上升子序列。

Code：

#include <algorithm>#include <iostream>#include <cstring>#include <string>#include <cstdio>#include <vector>#include <queue>#include <cmath>#include <map>#include <set>#define eps 1e-7#define LL long long#define pb push_back#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define Max(a,b) ((a)>(b)?(a):(b))#define Min(a,b) ((a)<(b)?(a):(b))using namespace std;const int maxn=250005;int dp[maxn];char str[maxn][20],tmp[20];char tar[]={'z','a','z'};void add(int id,int pos,char des){    strcpy(tmp,str[id]);    tmp[pos]=des;    while(str[id][pos]) tmp[pos+1]=str[id][pos],pos++;    tmp[pos+1]='\0';}void del(int id,int pos){    strcpy(tmp,str[id]);    pos++;    while(str[id][pos]) tmp[pos-1]=str[id][pos],pos++;    tmp[pos-1]='\0';}void change(int id,int pos,char des){    strcpy(tmp,str[id]);    tmp[pos]=des;    tmp[strlen(str[id])]='\0';}void edit(int id,int oper,int pos,char des){    switch(oper){        case 0:add(id,pos,des);break;        case 1:del(id,pos);break;        case 2:change(id,pos,des);break;    }}int bsearch(int l,int r){    while(l<=r){        int m=(l+r)>>1;        int x=strcmp(str[m],tmp) ;        if(x==0) return m;        else if(x<0) l=m+1;        else r=m-1;    }    return -1;}int main(){    int n=0;    while(scanf("%s",&str[n++])!=EOF) ;    int ans=1;    for(int i=0;i<n;i++){        dp[i]=1;        for(int op=0;op<3;op++){            for(int j=0;str[i][j];j++){                for(char c='a';c<=tar[op];c++){                    edit(i,op,j,c);                    int p=bsearch(0,i-1);                    if(p>=0&&dp[p]+1>dp[i]){                        dp[i]=dp[p]+1;                        if(dp[i]>ans) ans=dp[i];                    }                }            }        }    }    printf("%d\n",ans);    return 0;}