Codeforces gym 101343 J.Husam and the Broken Present 2【状压dp】

 2017 JUST Programming Contest 2.0

题目链接：Codeforces gym 101343 J.Husam and the Broken Present 2

J. Husam and the Broken Present 2

time limit per test：1.0 s

memory limit per test：256 MB

input：standard input

output：standard output

After you helped Husam and rebuilt his beautiful array a he became very happy. To avoid losing his array again, Husam made n copies from it, and distributed it to n of his friends. After that Husam became sure that he can rebuild the array a again if he lost it, so he destroyed the table t.

Today, Husam was looking for his array a, but he was not able to find it. Husam visited all his n friends to take a copy from the array. Unfortunately, all his friends thought that the length of the array a was very long, so instead of keeping the array itself, each friend i take a subarray (l_i, r_i) from the array and kept it in a safe place, and get rid of the rest of the array.

Now Husam has n subarrays from the array a, but he cannot remember the original array or even its length. Husam now needs your help again, he will give you the n subarrays, and your task is to build a new array a such that it contains all the given subarrays inside it as subarrays, and its length must be as minimal as possible. Can you?

Input

The first line contains an integer n (1 ≤ n ≤ 15), where n is the number of friends Husam has.

Then n lines follow, each line i begins with an integer m_i (1 ≤ m_i ≤ 100), where m_i is the length of the subarray the i^th friend has. Then m_i integers follow, representing the i^th subarray. All values x in the subarrays are in the range (1 ≤ x ≤ 10⁹).

Output

Print the minimal length of the new array a, such that a contains all the given subarrays in the input inside it as subarrays.

Examples

Input

3
2 1 2
4 3 4 5 6
3 2 3 4

Output

6

Input

5
3 4 7 5
4 7 9 2 5
3 7 5 2
4 5 1 4 7
4 9 2 5 1

Output

9

Note

A subarray of the array a is a sequence a_l, a_l + 1, ..., a_r for some integers (l, r) such that (1 ≤ l ≤ r ≤ n).

In the first test case the array a can be [1, 2, 3, 4, 5, 6]. Its length is as minimal as possible, and it contains all the the given subarrays in the input inside it as subarrays.

题意：构造一个序列包含所给的N个子序列(N≤15)，求构造序列的最短长度。

题解：

考虑状态压缩，有2^N种状态，设dp[i][j]表示状态为i，以第j个子序列结尾的最小长度；

状态转移：从以第j个子序列结尾的状态转移到以第k个子序列的状态：

　　dp[i|(1<<(k-1))][k]=min{dp[i][j]+a[k][0]-num[j][k]}   (其中，a[k][0]表示第k个子序列的长度，num[j][k]表示第j个子序列的后缀与第k个子序列的前缀重合部分的长度)

注意：先将被其他子序列包含的子序列删去......

【感觉自己现在一敲题就各种手残情况......】

代码：

#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
int a[16][101];
bool vis[16];       //vis[i]:第i个串是否被其他串包含
int num[16][16];
int dp[(1<<16)][16];//dp[i][j]:状态为i,以第j个子序列结尾的最小长度
int main() {
    int n, i, j, k, o, sum = 0, t, ans = 1e9;
    scanf("%d", &n);
    for(i = 1; i <= n; ++i) {
        scanf("%d", &a[i][0]);
        for(j = 1; j <= a[i][0]; ++j) scanf("%d", &a[i][j]);
    }
    for(i = 1; i <= n; ++i) if(!vis[i]){//删去被其他子序列包含的子序列
        for(j = 1; j <= n; ++j) {
            if(vis[j] || i == j || a[j][0] > a[i][0]) continue;

            for(k = 1; k <= a[i][0]-a[j][0]+1; ++k) {
                for(o = 1; o <= a[j][0]; ++o)
                    if(a[i][k+o-1] != a[j][o]) break;
                if(o == a[j][0]+1) {vis[j] = true; break;}
            }
        }
    }
    for(i = 1; i <= n; ++i) if(!vis[i]) {
        for(sum++,j=0; j<=a[i][0]; ++j) a[sum][j] = a[i][j];
    }
    n = sum;
    for(i = 1; i <= n; ++i) {//计算num[i][j]
        for(j = 1; j <= n; ++j) {
            if(i == j) continue;
            for(k = 1; k <= a[i][0]; ++k) {//枚举重合长度
                for(t=1, o=a[i][0]-k+1; o <= a[i][0]; ++o)
                    if(a[i][o] != a[j][t++]) break;
                if(o == a[i][0]+1) num[i][j] = k;
            }
        }
    }
    sum = (1<<n)-1; //状态总数
    memset(dp, inf, sizeof(dp));
    for(i = 1; i <= n; ++i) dp[1<<(i-1)][i] = a[i][0];
    for(i = 1; i <= sum; ++i) {//计算dp[i][j]
        for(j = 1; j <= n; ++j) {
            if(!((1<<(j-1))&i)) continue;
            for(k = 1; k <= n; ++k) {
                if((1<<(k-1))&i) continue;
                dp[i|(1<<(k-1))][k]=min(dp[i|(1<<(k-1))][k],dp[i][j]+a[k][0]-num[j][k]);
            }
        }
    }
    for(i = 1; i <= n; ++i) ans = min(ans, dp[sum][i]);
    printf("%d\n", ans);
    return 0;
}