poj 2817

 

WordStack

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 2088 Accepted: 696

Description

As editor of a small-town newspaper, you know that a substantial number of your readers enjoy the daily word games that you publish, but that some are getting tired of the conventional crossword puzzles and word jumbles that you have been buying for years. You decide to try your hand at devising a new puzzle of your own. 

Given a collection of N words, find an arrangement of the words that divides them among N lines, padding them with leading spaces to maximize the number of non-space characters that are the same as the character immediately above them on the preceding line. Your score for this game is that number.

Input

Input data will consist of one or more test sets. 

The first line of each set will be an integer N (1 <= N <= 10) giving the number of words in the test case. The following N lines will contain the words, one word per line. Each word will be made up of the characters 'a' to 'z' and will be between 1 and 10 characters long (inclusive). 

End of input will be indicated by a non-positive value for N .

Output

Your program should output a single line containing the maximum possible score for this test case, printed with no leading or trailing spaces.

Sample Input

5 abc bcd cde aaa bfcde 0

Sample Output

8

Hint

Note: One possible arrangement yielding this score is: 

aaa 
abc 
 bcd
  cde 
bfcde

Source

Mid-Atlantic 2005

分析：状态压缩DP，把取单词的情况用二进制表示，容易得到状态转移方程

    f[i][s|pow[i]]=max(f[i][s|pow[i]],f[j][s]+w[i][j]),其中 s为当前状态,，pow[i]=1<<(i-1);pow[i]&s==0,pow[j]&s>0,w[i][j]为单词i，j相邻的值

还是好挫地wrong了一次，最近实在是没状态啊~~~‍

找来一组数据：

 

10

abc

bc

abcd

bbc

bcd

dbc

a

bacd

ebdac

ac

答案：19

 

代码：

#include<cstdio>#include<cstring>using namespace std;int p[11]={1,2,4,8,16,32,64,128,256,512,1024},w[11][11],f[11][1025];char word[11][11];inline int max(int a,int b){return a>b?a:b;}void get(int x,int y){ int i,j,k,s; w[x][y]=0; for(i=0;word[x][i]!='/0';++i) for(j=0;word[y][j]!='/0';++j) { k=s=0; while(word[x][i+k]!='/0'&&word[y][j+k]!='/0') { if(word[x][i+k]==word[y][j+k])++s; ++k; } if(s>w[x][y])w[x][y]=s; } w[y][x]=w[x][y];}int main(){ int i,j,k,n,ans; while(scanf("%d",&n),n) { memset(f,0,sizeof(f)); for(i=0;i<n;++i)scanf("%s",word[i]); for(i=0;i<n;++i) for(j=i+1;j<n;++j)get(i,j); for(k=0;k<p[n];++k) for(i=0;i<n;++i) if((k&p[i])==0) for(j=0;j<n;++j) if(k&p[j])f[i][k|p[i]]=max(f[i][k|p[i]],f[j][k]+w[i][j]); for(i=ans=0;i<n;++i)ans=max(ans,f[i][p[n]-1]); printf("%d/n",ans); } return 0;}