poj 2817 状态DP

WordStack
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 3248 Accepted: 1171
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

#include <stdio.h>#include <string.h>#include <string>using namespace std;int dp[1<<(12)][12];int max(int a, int b){if(a > b)return a;return b;}int main(){char str[12][12];int n;int len[12][12]; //len[i][j]表示第i个字符串和第j个字符串的最大公共字符长度while((scanf("%d", &n) != EOF) && (n > 0)){memset(len, 0, sizeof(len));int i, j, k, l, count, maxx;for(i = 0; i < n; i++){scanf("%s", str[i]);}for(i = 0; i < n; i++){for(j = 0; j < n; j++) {if(i != j){int len1 = strlen(str[i]);int len2 = strlen(str[j]);maxx = -1;for(k = 0; k < len1; k++){count = 0;for(l = 0; (l < len2) && (l+k < len1); l++)//for(l=0;(l<len1 && l+k < len1); l++) if(str[i][l] == str[j][l+k]) count++{if(str[i][l+k] == str[j][l])count++;}if(count > maxx)maxx = count;}if(maxx > len[i][j])len[i][j] = len[j][i] = maxx;}}}memset(dp, 0, sizeof(dp));for(i = 0; i < (1<<n); i++){for(j = 0; j < n; j++){if(i&(1<<j)){for(k = 0; k < n; k++){if(!(i&(1<<k))){dp[i|(1<<k)][k] = max(dp[i|(1<<k)][k], dp[i][j]+len[j][k]);}}}}}maxx = -1;for(i = 0; i < (1<<n); i++){for(j = 0; j < n; j++){if(dp[i][j] > maxx)maxx = dp[i][j];}}printf("%d\n", maxx);}return 0;}