Mix and Build(简单DP)

Mix and Build
Time Limit: 5000MS Memory Limit: 65536K
Total Submissions: 3936 Accepted: 1203
Case Time Limit: 2000MS Special Judge

Description
In this problem, you are given a list of words (sequence of lower case letters). From this list, find the longest chain of words w1, …, wn such that wi is a mixed extension of wi-1. A word A is a mixed extension of another word B if A can be formed by adding one letter to B and permuting the result. For example, “ab”, “bar”, “crab”, “cobra”, and “carbon” form a chain of length 5.

Input
The input contains at least two, but no more than 10000 lines. Each line contains a word. The length of each word is at least 1 and no more than 20. All words in the input are distinct.

Output
Write the longest chain that can be constructed from the given words. Output each word in the chain on a separate line, starting from the first one. If there are multiple longest chains, any longest chain is acceptable.

Sample Input

ab
arc
arco
bar
bran
carbon
carbons
cobra
crab
crayon
narc

Sample Output

ab
bar
crab
cobra
carbon
carbons

Source
Rocky Mountain 2004
简单的Dp,找最长字符串链,使后一个比前一个的长度大一,并且只有一个字符不同

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <map>
#include <algorithm>
#define INF 0x3f3f3f3f
using namespace std;

typedef unsigned long long LL;

const int MAX = 1e5+10;

struct node
{
    char str[25];
    int Hash[26];
    int len;
    int Dp;
    int pre;
    void init()//初始化
    {
        memset(Hash,0,sizeof(Hash));
        len=0;
        Dp=1;
        pre=-1;
    }
    void HASH()//哈希字符
    {
        len=strlen(str);
        for(int i=0;i<len;i++)
        {
            Hash[str[i]-'a']++;
        }
    }
    void Output()
    {
        printf("%s\n",str);
    }
}Ch[11000];

bool cmp(node a,node b)//按照长度进行排序
{
    return a.len<b.len;
}

void DFS(int s)//输出
{
    if(s==-1)
    {
        return ;
    }
    DFS(Ch[s].pre);
    Ch[s].Output();
}

int main()
{
   // freopen("input.txt","r",stdin);
    for(int i=0;i<10001;i++)
    {
        Ch[i].init();
    }
    int top=0;
    while(~scanf("%s",Ch[top].str))
    {
        Ch[top].HASH();
        top++;
    }
    sort(Ch,Ch+top,cmp);
    for(int i=0;i<top;i++)
    {
        for(int j=i-1;j>=0;j--)
        {
            if(Ch[j].len==Ch[i].len)
            {
                continue;
            }
            if(Ch[j].len==Ch[i].len-1)
            {
                int ans=0;
                for(int k=0;k<26;k++)
                {
                    if(Ch[i].Hash[k]!=Ch[j].Hash[k])
                    {
                        ans++;
                    }
                    if(ans>2)
                    {
                        break;
                    }
                }
                if(ans<2)//如果不同的字符大于两个则不符合
                {
                    if(Ch[i].Dp<Ch[j].Dp+1)
                    {
                        Ch[i].Dp=Ch[j].Dp+1;
                        Ch[i].pre=j;
                    }
                }
            }
            else
            {
                break;
            }
        }
    }
    int Max=0,ans;
    for(int i=0;i<top;i++)
    {
        if(Max<Ch[i].Dp)
        {
            Max=Ch[i].Dp;
            ans=i;
        }
    }
    DFS(ans);
    return 0;
}