1195: [HNOI2006]最短母串

思路：好像以前谁问过我这题。。。  状个压就好啦， 把包含在其他串中的字符串删掉， 预处理除每两个字符串之间的关系，

dp[ state ][ i ] 表示在state的状态下， 最后一个字符串是第i个的最优解， 字典序最小的话暴力把转移过程中的字符串存下来

就好啦。

 #include<bits/stdc++.h>
 #define LL long long
 #define fi first
 #define se second
 #define mk make_pair
 #define pii pair<int,int>
 #define piii pair<int, pair<int,int>>

 using namespace std;

 const int N=+;
 const int M=1e4+;
 const int inf=0x3f3f3f3f;
 const LL INF=0x3f3f3f3f3f3f3f3f;
 const int mod=1e9 + ;

 int n, top, up, cost[][], len[], ret[], dp1[ << ][];
 string s[], t[], dp2[ << ][];

 int cal(string &a, string &b, int len1, int len2) {
     for(int i = max(, len2 - len1); i < len2; i++) {
         bool flag = true;
         for(int j = i, k = ; j < len2; j++, k++) {
             if(b[j] != a[k]) {
                 flag = false;
                 break;
             }
         }
         if(flag) {
             return len1 - (len2 - i);
         }
     }
     return len1;
 }

 bool check(string &t) {
     for(int i = ; i < top; i++) {
         for(int j = ; j < s[i].size(); j++) {
             string ret = s[i].substr(j, t.size());
             if(ret == t) return false;
         }
     }
     return true;
 }

 bool cmp(const string &a, const string &b) {
     return a.size() > b.size();
 }
 int main() {
     memset(dp1, inf, sizeof(dp1));
     scanf("%d", &n);
     for(int i = ; i < n; i++) {
         cin >> t[i];
     }

     sort(t, t + n, cmp);

     for(int i = ; i < n; i++) {
         if(check(t[i])) {
             s[top++] = t[i];
         }
     }

     n = top; up =  << n;

     for(int i = ; i < n; i++)
         len[i] = s[i].size();

     for(int i = ; i < n; i++) {
         for(int j = ; j < n; j++) {
             if(i == j) continue;
             cost[i][j] = cal(s[i], s[j], len[i], len[j]);
         }
     }

     for(int i = ; i < n; i++) {
         dp1[ << i][i] = len[i];
         dp2[ << i][i] = s[i];
     }

     for(int state = ; state < up; state++) {
         for(int i = ; i < n; i++) {
             if(dp1[state][i] == inf)
                 continue;
             for(int j = ; j < n; j++) {
                 if((state >> j) & ) continue;
                 if(dp1[state ^ ( << j)][j] > dp1[state][i] + cost[j][i]) {
                     dp1[state ^ ( << j)][j] = dp1[state][i] + cost[j][i];
                     dp2[state ^ ( << j)][j] = dp2[state][i] + s[j].substr(len[j] - cost[j][i], cost[j][i]);
                 } else if(dp1[state ^ ( << j)][j] == dp1[state][i] + cost[j][i]) {
                     string ret = dp2[state][i] + s[j].substr(len[j] - cost[j][i], cost[j][i]);
                     if(ret < dp2[state ^ ( << j)][j])
                         dp2[state ^ ( << j)][j] = dp2[state][i] + s[j].substr(len[j] - cost[j][i], cost[j][i]);
                 }
             }
         }
     }

     int id = ;

     for(int i = ; i < n; i++) {
         if(dp1[up - ][i] < dp1[up - ][id]) {
             id = i;
         } else if(dp1[up - ][i] == dp1[up - ][id] && dp2[up - ][i] < dp2[up - ][id]) {
             id = i;
         }
     }

     cout << dp2[up - ][id] << endl;
     return ;
 }
 /*
 */