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=100+7;
const int M=1e4+7;
const int inf=0x3f3f3f3f;
const LL INF=0x3f3f3f3f3f3f3f3f;
const int mod=1e9 + 7;

int n, top, up, cost[12][12], len[12], ret[12], dp1[1 << 12][12];
string s[12], t[12], dp2[1 << 12][12];

int cal(string &a, string &b, int len1, int len2) {
    for(int i = max(0, len2 - len1); i < len2; i++) {
        bool flag = true;
        for(int j = i, k = 0; 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 = 0; i < top; i++) {
        for(int j = 0; 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 = 0; i < n; i++) {
        cin >> t[i];
    }

    sort(t, t + n, cmp);

    for(int i = 0; i < n; i++) {
        if(check(t[i])) {
            s[top++] = t[i];
        }
    }

    n = top; up = 1 << n;

    for(int i = 0; i < n; i++)
        len[i] = s[i].size();

    for(int i = 0; i < n; i++) {
        for(int j = 0; j < n; j++) {
            if(i == j) continue;
            cost[i][j] = cal(s[i], s[j], len[i], len[j]);
        }
    }

    for(int i = 0; i < n; i++) {
        dp1[1 << i][i] = len[i];
        dp2[1 << i][i] = s[i];
    }

    for(int state = 1; state < up; state++) {
        for(int i = 0; i < n; i++) {
            if(dp1[state][i] == inf)
                continue;
            for(int j = 0; j < n; j++) {
                if((state >> j) & 1) continue;
                if(dp1[state ^ (1 << j)][j] > dp1[state][i] + cost[j][i]) {
                    dp1[state ^ (1 << j)][j] = dp1[state][i] + cost[j][i];
                    dp2[state ^ (1 << j)][j] = dp2[state][i] + s[j].substr(len[j] - cost[j][i], cost[j][i]);
                } else if(dp1[state ^ (1 << 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 ^ (1 << j)][j])
                        dp2[state ^ (1 << j)][j] = dp2[state][i] + s[j].substr(len[j] - cost[j][i], cost[j][i]);
                }
            }
        }
    }

    int id = 0;

    for(int i = 1; i < n; i++) {
        if(dp1[up - 1][i] < dp1[up - 1][id]) {
            id = i;
        } else if(dp1[up - 1][i] == dp1[up - 1][id] && dp2[up - 1][i] < dp2[up - 1][id]) {
            id = i;
        }
    }

    cout << dp2[up - 1][id] << endl;
    return 0;
}
/*
*/