动态规划之Edit Step Ladders

题目：

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. For a given dictionary, you are to compute the length of the longest edit step ladder.
Input
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
cat

dig

dog

fig

fin

fine

fog

log

wine
Sample Output
5

动态规划很好想，就是在判断的时候先把一个词所有可能的变换找出来，然后由于单词有序可以二分查找，否则一个一个去比会超时...这道题因为变换的时候末尾没加\0,wa了超多次..郁闷死我了....大致思路是这样，其中还有一些点应该还可以优化....啊我去，先不管了...

#include<stdio.h>#include<string.h>char word[25000][20];int dp[25000];int find(int r,char target[]){    int l,mid,flag;    l=0;    while(l<=r){        mid=(l+r)/2;        flag=strcmp(word[mid],target);        if(flag==0)            return mid;        else if(flag>0)            r=mid-1;        else            l=mid+1;    }    return -1;}int change(int pos,char t[],int i,char c){    strcpy(t,word[i]);    t[pos]=c;    return find(i-1,t);}int del(int pos,char t[],int i){    int k,j=0;    for(k=0;k<strlen(word[i]);k++){        if(pos==k)            continue;        t[j]=word[i][k];        j++;    }    t[j]='\0';    return find(i-1,t);}int add(int pos,char t[],int j,char c){    int i;    strcpy(t,word[j]);    t[pos]=c;    for(i=pos;i<strlen(word[j]);i++)        t[i+1]=word[j][i];    t[strlen(word[j])+1]='\0';    return find(j-1,t);}int main(){    int n=0;    int i,pos,index,tmp,ans;    char c;    char t[30];    while(scanf("%s",word[n])!=EOF)        n++;    dp[0]=ans=1;    for(i=1;i<n;i++){        tmp=0;        for(c='a';c<'a'+26;c++){            for(pos=0;pos<=strlen(word[i]);pos++){                index=add(pos,t,i,c);                if(index>=0&&dp[index]>tmp)                    tmp=dp[index];            }            for(pos=0;pos<strlen(word[i]);pos++){                index=change(pos,t,i,c);                if(index>=0&&dp[index]>tmp)                    tmp=dp[index];            }        }        for(pos=0;pos<strlen(word[i]);pos++){            index=del(pos,t,i);            if(index>=0&&dp[index]>tmp)                tmp=dp[index];        }        dp[i]=tmp+1;        if(dp[i]>ans)            ans=dp[i];    }    printf("%d\n",ans);    return 0;}