poj2817（状态压缩dp基础题）

WordStack

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3435		Accepted: 1254

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

题意：给n个字符串，每两个字符串之间匹配有一个数值（匹配只能是在某一个字符串前面加空格来匹配，不是最长公共子串），然后要使整体的匹配数值最大，求最大数值。主要还是要看排列顺序是什么样。

思路：由于数据只给到10了所以直接状态压缩，因为是两两之间有个匹配值所以要预处理两两之间的匹配值然后进行状态压缩dp就好。

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<queue>
#include<stack>
using namespace std;
typedef long long ll;
char a[11][11];
int dp[1<<10][11];
int w[11][11];
void get(int x,int y)
{
    w[x][y]=w[y][x]=0;
    int l1=strlen(a[x]),l2=strlen(a[y]),s=0;
    for(int i=0; i<l1; i++)
        for(int j=0; j<l2; j++)
        {
            int k=0;
            s=0;
            while(i+k<l1&&j+k<l2)
            {
                if(a[x][i+k]==a[y][j+k])
                    s++;
                k++;
            }
            if(s>w[x][y])w[x][y]=w[y][x]=s;
        }
}

int main()
{
    int n;
    while(~scanf("%d",&n)&&n)
    {
        for(int i=0; i<n; i++)
            scanf("%s",a[i]);
        for(int i=0; i<n; i++)
            for(int j=i+1; j<n; j++)
                get(i,j);
        memset(dp,0,sizeof(dp));
        for(int i=0; i<(1<<n); i++)
            for(int j=0; j<n; j++)
                if(i&(1<<j))
                    for(int k=0; k<n; k++)
                        if(!(i&(1<<k)))
                            if(dp[i][j]+w[j][k]>dp[i|(1<<k)][k])
                                dp[i|(1<<k)][k]=dp[i][j]+w[j][k];
        int ans=0;
        for(int i=0; i<n; i++)
            ans=max(ans,dp[(1<<n)-1][i]);
        cout<<ans<<endl;
    }
    return 0;
}