Uva--10029（动规，map，记忆化搜索）

2014-08-04 22:52:19

Problem C: Edit Step Ladders

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, ... w_n such that the transformation from w_i to w_i+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

借鉴：http://blog.csdn.net/wiking__acm/article/details/24382243
思路：字符版LIS，25000个字符，如果用朴素O(n^2)算法会超时（如果Uva数据正常的话），固考虑用dp记忆化搜索，也就是DAG上求最长路的问题了，关键在于图的连边上怎么处理呢。
这里借鉴了别人博客，有add，delete，change三种转换方式，其中add和delete是等价的（串A增一个字符得到串B，那么串B删一个字符必能得到串A）。这里用‘？’字符来插入串中某个位置（‘？’表示可为任意字母），或代替串中某个字符来达到add和change的效果，再把转换后的串进行匹配。这样就可以建图了，记忆化搜索即可。

/*************************************************************************
    > File Name: j.cpp
    > Author: Nature
    > Mail: 564374850@qq.com 
    > Created Time: Mon 04 Aug 2014 02:42:06 PM CST
************************************************************************/

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <map>
#include <string>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 25000;

map<string,int> m;
vector<int> g[maxn + 5];
string s[maxn + 5];
int cnt;

string Chan(const string &st,int pos,int len){
    string tem;
    for(int i = 0; i < len; ++i){
        if(i == pos) tem.push_back('?');
        else tem.push_back(st[i]);
    }
    return tem;
}

string Add(const string &st,int pos,int len){
    string tem;
    for(int i = 0; i < len; ++i){
        if(i == pos) tem.push_back('?');
        tem.push_back(st[i]);
    }
    if(pos == len)
        tem.push_back('?');
    return tem;
}

int dp[maxn + 5];

int Solve(int p,int num){
    if(dp[p] != -1)
        return dp[p];
    int tmax = 0;
    for(int i = 0; i < g[p].size(); ++i){
        tmax = max(tmax,Solve(g[p][i],num + 1));
    }
    return dp[p] = 1 + tmax; //记忆化搜索核心，tmax是后继子情况最优解
}

int main(){
    cnt = 0;
    while(cin >> s[cnt])
        ++cnt;
    for(int i = cnt - 1; i >= 0; --i){
        int l = s[i].size();
        for(int p = 0; p < l; ++p){
            string tChan = Chan(s[i],p,l);
            if(m.count(tChan) > 0)
                g[m[tChan]].push_back(i);
            m[tChan] = i;
        }
        for(int p = 0; p <= l; ++p){
            string tAdd = Add(s[i],p,l);
            if(m.count(tAdd) > 0)
                g[m[tAdd]].push_back(i);
            m[tAdd] = i;
        }
    }
    memset(dp,-1,sizeof(dp));
    int ansmax = 0;
    for(int i = cnt - 1; i >= 0; --i)
        ansmax = max(ansmax,Solve(i,1));
    printf("%d\n",ansmax);
    return 0;
}