POJ1033-Defragment

Defragment

Time Limit: 2000MS		Memory Limit: 10000K
Total Submissions: 4298		Accepted: 1478
Case Time Limit: 1000MS		Special Judge

Description

You are taking part in the development of a "New Generation" operating system and the NG file system. In this file system all disk space is divided into N clusters of the equal sizes, numbered by integers from 1 to N. Each file occupies one or more clusters in arbitrary areas of the disk. All clusters that are not occupied by files are considered to be free. A file can be read from the disk in the fastest way, if all its clusters are situated in the successive disk clusters in the natural order. 
Rotation of the disk with constant speed implies that various amounts of time are needed for accessing its clusters. Therefore, reading of clusters located near the beginning of the disk performs faster than reading of the ones located near its ending. Thus, all files are numbered beforehand by integers from 1 to K in the order of descending frequency of access. Under the optimal placing of the files on the disk the file number 1 will occupy clusters 1, 2, ..., S1, the file number 2 will occupy clusters S1+1, S1+2, ..., S1+S2 and so on (here Si is the number of clusters which the i-th file occupies). 
In order to place the files on the disk in the optimal way cluster-moving operations are executed. One cluster-moving operation includes reading of one occupied cluster from the disk to the memory and writing its contents to some free cluster. After that the first of them is declared free, and the second one is declared occupied. 
Your goal is to place the files on the disk in the optimal way by executing the minimal possible number of cluster-moving operations. 

Input

The first line of the input file contains two integers N and K separated by a space(1 <= K < N <= 10000).Then K lines follow, each of them describes one file. The description of the i-th file starts with the integer Si that represents the number of clusters in the i-th file (1 <= Si < N). Then Si integers follow separated by spaces, which indicate the cluster numbers of this file on the disk in the natural order. 
All cluster numbers in the input file are different and there is always at least one free cluster on the disk. 

Output

Your program should write to the output file any sequence of cluster-moving operations that are needed in order to place the files on the disk in the optimal way. Two integers Pj and Qj separated by a single space should represent each cluster-moving operation. Pj gives the cluster number that the data should be moved FROM and Qj gives the cluster number that this data should be moved TO. 
The number of cluster-moving operations executed should be as small as possible. If the files on the disk are already placed in the optimal way the output should contain only the string "No optimization needed". 

Sample Input

20 3
4 2 3 11 12
1 7
3 18 5 10

Sample Output

2 1
3 2
11 3
12 4
18 6
10 8
5 20
7 5
20 7

Source

Northeastern Europe 1998

题意：要实现磁盘碎片整理的功能。磁盘分为N个簇，一个文件可以占用K个簇，(1 <= K < N <= 10000)，给出各个文件的占用磁盘的情况，也就是一个文件占用了哪些簇，进行碎片整理，把这些簇按顺序整理到磁盘的最顶部

样例：

文件1：2 3 11 12，占用了4个簇，编号为1-4。
文件2：7，占用了1个簇，编号为5。
文件3：18 5 10，占用了3个簇，编号为6-8。

初始状态是这样的，0表示未占用：
簇号：　　1  2  3  4  5  6  7  8  9  10  11  12  13  14  15  16  17  18
逻辑编号：0  1  2  0  7  0  5  0  0   8   3   4   0   0   0   0   0   6 

一共整理到最后，磁盘的情况最后是这样的：
簇号：　　1  2  3  4  5  6  7  8  9  10  11  12  13  14  15  16  17  18
逻辑编号：1  2  3  4  5  6  7  8  0   0   0   0   0   0   0   0   0   0 

写一个程序得到整理好碎片最少需要多少步操作，并把这些操作打印出来。比如第1个簇的内容放到第2个簇，打印出1 2。把一个簇的内容放到另个一个簇中，算是一步操作。
这里是Special Judge，只要答案符合要求就行了，不必和SAMPLE中的OUTPUT一样也可以AC。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>

using namespace std;

int a[10005];
int n,k;
int cur;
int ring;

void Swap(int x,int y)
{
    a[y]=a[x];
    a[x]=0;
    printf("%d %d\n",x,y);
}

void dfs(int x)
{
    if(!a[x]) return ;
    else if(a[x]==cur)
    {
        for(int i=n; i>=1; i--)
        {
            if(!a[i])
            {
                ring=i;
                Swap(x,i);
                break;
            }
        }
    }
    else
    {
        dfs(a[x]);
        Swap(x,a[x]);
    }
}

void solve(int x)
{
    cur=x;
    ring=0;
    dfs(x);
    if(ring)
        Swap(ring,x);
}

int main()
{
    while(~scanf("%d %d",&n,&k))
    {
        int num=1,x;
        memset(a,0,sizeof a);
        for(int i=1; i<=k; i++)
        {
            int t;
            scanf("%d",&t);
            for(int j=1; j<=t; j++)
            {
                scanf("%d",&x);
                a[x]=num;
                num++;
            }
        }
        bool flag=0;
        for(int i=1; i<=n; i++)
        {
            if(a[i]!=i)
            {
                if(!a[i]) continue;
                flag=1;
                solve(i);
            }
        }
        if(!flag) printf("No optimization needed\n");
    }
    return 0;
}