uva 10570 Meeting with Aliens

最新推荐文章于 2020-12-28 11:04:19 发布

原创最新推荐文章于 2020-12-28 11:04:19 发布 · 1.3k 阅读

0 ·

CC 4.0 BY-SA版权

杂题专栏收录该内容

21 篇文章

订阅专栏

解决外星人在圆桌会议上座位顺序混乱的问题，通过最少的交换次数使1到N的编号顺序正确，涉及算法分析和实现。

Problem D
Meeting with Aliens
Input: Standard Input

Output: Standard Output

Time Limit: 3 Seconds

The aliens are in an important meeting just before landing on the earth. All the aliens sit around a round table during the meeting. Aliens are numbered sequentially from 1 to N. It's a precondition of the meeting that i'th alien will sit between (i-1)'th and (i+1)'th alien. 1st alien will sit between 2nd and N'th alien.

Though the ordering of aliens are fixed but their positions are not fixed. In the above figure two valid sitting arrangements of eight aliens are shown. Right before the start of the meeting the aliens sometimes face a common problem of not maintaining the proper order. This occurs as no alien has a fixed position. Two maintain the proper order, two aliens can exchange their positions. The aliens want to know the minimum number of exchange operations necessary to fix the order.

Input

Input will start with a positive integer, N (3<=N<=500) the number of aliens. In next few lines there will be N distinct integers from 1 to N indicating the current ordering of aliens. Input is terminated by a case where N=0. This case should not be processed. There will be not more than 100 datasets.

Output

For each set of input print the minimum exchange operations required to fix the ordering of aliens.

Sample Input Output for Sample Input

1 2 3 4

4 3 2 1

2 3 1 4

Problemsetter: Md. Kamruzzaman, Member of Elite Problemsetters' Panel

Special thanks to Derek Kisman

求最少交换次数，使得1~n排列有序。找出环的数目，答案就是n-环数。eg （13254）可分为（1）、（23）、（54）三个环。

#include<cstdio>
#include<map>
#include<queue>
#include<cstring>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
const int maxn = 1000 + 5;
const int INF = 1000000000;
typedef long long LL;
typedef pair<LL, int> P;

int a[maxn];
int vis[maxn];
int n;

int solve(int l, int r){
    memset(vis, 0, sizeof  vis);
    int cnt = 0;
    for(int i = l;i <= r;i++){
        if(vis[i] == 1)
            continue;
        cnt++;
        int start = i, now = a[i]+l-1;
        while(now != start){
            vis[now] = 1;
            now = a[now]+l-1;
        }
    }
    return n-cnt;
}

int main(){
    while(scanf("%d", &n)){
        if(n == 0) break;
        for(int i = 0;i < n;i++){
            scanf("%d", &a[i]);
            a[i+n] = a[i];
        }
        int ans = INF;
        for(int i = 0;i < n;i++)
            ans = min(ans, solve(i, i+n-1));
        reverse(a, a+n);
        for(int i = 0;i < n;i++)
            a[i+n] = a[i];
        for(int i = 0;i < n;i++)
            ans = min(ans, solve(i, i+n-1));
        printf("%d\n", ans);
    }
    return 0;
}