CF 53D

题意：

给定两个数组，求由第一个数组到第二个数组的操作方法数 和 步骤（只能相邻间交换）

暴力+模拟

/*

*/
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
#include<queue>
//#include<map>
#include<math.h>
using namespace std;
typedef long long ll;
//typedef __int64 int64;
const int maxn = 305;
const int inf = 0x7fffffff;
const double pi=acos(-1.0);
int a[ maxn ],b[ maxn ];
int vis_b[ maxn ];
struct node{
    int x,y;
};
queue<node>step[ maxn ];
void init(){
    for( int i=0;i<maxn;i++ ){
        while( !step[ i ].empty() )
            step[ i ].pop();
    }
    memset( vis_b,0,sizeof( vis_b ) );
}
int main(){
    int n;
    while( scanf("%d",&n)!=EOF ){
        for( int i=0;i<n;i++ )
            scanf("%d",&a[ i ]);
        for( int i=0;i<n;i++ )
            scanf("%d",&b[ i ]);
        init();
        int cnt=0;
        node now;
        for( int i=0;i<n;i++ ){
            for( int j=0;j<n;j++ ){
                if( vis_b[ j ]==0&&a[ i ]==b[ j ] ){
                    vis_b[ i ]=1;
                    if( i==j ){
                        //vis_b[ j ]=1;
                        break;
                    }
                    else if( i<j ){
                        //vis_b[ j ]=1;
                        now.x=j-1,now.y=j;
                        step[ cnt ].push( now );
                        swap( b[ now.x ],b[ now.y ] );
                        while( 1 ){
                            if( now.x==i ) break;
                            now.x--;
                            now.y--;
                            step[ cnt ].push( now );
                            swap( b[ now.x ],b[ now.y ] );
                        }
                        cnt++;
                        break;
                    }
                    else if( i>j ){
                        //vis_b[ j ]=1;
                        now.x=j,now.y=j+1;
                        step[ cnt ].push( now );
                        swap( b[ now.x ],b[ now.y ] );
                        while( 1 ){
                            if( now.y==i ) break;
                            now.x++;
                            now.y++;
                            step[ cnt ].push( now );
                            swap( b[ now.x ],b[ now.y ] );
                        }
                        cnt++;
                        break;
                    }
                }
            }
        }
        int ans=0;
        for( int i=0;i<cnt;i++ )
            ans+=( step[ i ].size() );
        printf("%d\n",ans);
        for( int i=0;i<cnt;i++ ){
            while( !step[ i ].empty() ){
                now=step[ i ].front();
                printf("%d %d\n",1+now.x,1+now.y);
                step[ i ].pop();
            } 
        }
    }
    return 0;
}