codefores New Year Cards(模拟题)

B - New Year Cards

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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Description

As meticulous Gerald sets the table, Alexander finished another post on Codeforces and begins to respond to New Year greetings from friends. Alexander has n friends, and each of them sends to Alexander exactly one e-card. Let us number his friends by numbers from 1to n in the order in which they send the cards. Let's introduce the same numbering for the cards, that is, according to the numbering the i-th friend sent to Alexander a card number i.

Alexander also sends cards to friends, but he doesn't look for the new cards on the Net. He simply uses the cards previously sent to him (sometimes, however, he does need to add some crucial details). Initially Alexander doesn't have any cards. Alexander always follows the two rules:

1. He will never send to a firend a card that this friend has sent to him.
2. Among the other cards available to him at the moment, Alexander always chooses one that Alexander himself likes most.

Alexander plans to send to each friend exactly one card. Of course, Alexander can send the same card multiple times.

Alexander and each his friend has the list of preferences, which is a permutation of integers from 1 to n. The first number in the list is the number of the favorite card, the second number shows the second favorite, and so on, the last number shows the least favorite card.

Your task is to find a schedule of sending cards for Alexander. Determine at which moments of time Alexander must send cards to his friends, to please each of them as much as possible. In other words, so that as a result of applying two Alexander's rules, each friend receives the card that is preferred for him as much as possible.

Note that Alexander doesn't choose freely what card to send, but he always strictly follows the two rules.

Input

The first line contains an integer n (2 ≤ n ≤ 300) — the number of Alexander's friends, equal to the number of cards. Next n lines contain his friends' preference lists. Each list consists of n different integers from 1 to n. The last line contains Alexander's preference list in the same format.

Output

Print n space-separated numbers: the i-th number should be the number of the friend, whose card Alexander receives right before he should send a card to the i-th friend. If there are several solutions, print any of them.

Sample Input

Input

41 2 3 44 1 3 24 3 1 23 4 2 13 1 2 4

Output

2 1 1 4

Hint

In the sample, the algorithm of actions Alexander and his friends perform is as follows:

1. Alexander receives card 1 from the first friend.
2. Alexander sends the card he has received (at the moment he only has one card, and therefore it is the most preferable for him) to friends with the numbers 2 and 3.
3. Alexander receives card 2 from the second friend, now he has two cards — 1 and 2.
4. Alexander sends a card to the first friend. Despite the fact that Alexander likes card 1 more, he sends card 2 as he cannot send a friend the card sent by that very friend.
5. Alexander receives card 3 from the third friend.
6. Alexander receives card 4 from the fourth friend.
7. Among the cards Alexander has number 3 is his favorite and he sends it to the fourth friend.

Note that Alexander can send cards to multiple friends at a time (in this case the second and the third one). Alexander can send card 3 to the fourth friend after he receives the third card or after he receives the fourth card (both variants are correct).

题意:Alexander 与他的朋友互发贺卡，朋友依次从第一个编号到最后一个编号向Alexander 发贺卡，第i个朋友发的贺卡为i，而Alexander 发贺卡有限制:

1.Alexander 最开始是没有贺卡的

2.Alexander 收到某人的贺卡，不能将这个贺卡发回去（即朋友i不能收自己本身发出去的贺卡i）。

3.Alexander 每次发贺卡的时候只能发当前拥有的贺卡中，自己最喜欢的贺卡给朋友。

4.Alexander可以同时发多张贺卡给不同的朋友

 

如此我们可以运用数值覆盖的思路，针对当前拥有的所有卡片，将最好的贺卡都发给所有朋友(注意第二个限制)。而每一个朋友他们自己根据自己的喜爱级别进行选择就可以了，具体讲解请看代码中的注解。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int MAXN = 3e2 + 5;int n;int O[MAXN][MAXN], Alex[MAXN];bool vis[MAXN];int r[MAXN];int main() {while(~scanf("%d", &n)) {for(int i = 1; i <= n; i ++) {for(int j = 1; j <= n; j ++) {scanf("%d", &O[i][j]);}}int x;for(int i = 1; i <= n; i ++) {scanf("%d", &x);Alex[x] = i;//用于比较从而得到当前拥有的最好的贺卡}for(int i = 1; i <= n; i ++) {memset(vis, false, sizeof(vis));//用于标记拥有什么贺卡int o = -1;//初始化当前没有贺卡for(int j = 1; j <= n; j ++) {if(i == j) continue;//本身发的贺卡不能收取if(o != -1 && Alex[j] > Alex[o]) continue;//如果当前朋友发过来的贺卡并不比当前Alexander拥有的最好贺卡的喜爱级别更高也抛弃o = j;//当前最好的卡片vis[j] = true;//标记拥有这个卡片}for(int j = 1; j <= n; j ++) {if(vis[O[i][j]]) {//由于O[i][j]本身就具有优先性，即从1...n他们的喜爱级别是不断减少的，所以，当检测到第一个的时候就跳出循环r[i] = O[i][j];break;}}}for(int i = 1; i <= n; i ++) {printf("%d%c", r[i], i == n ? '\n' : ' ');}}return 0;}