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

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 1 /*

 2 

 3 */

 4 #include<stdio.h>

 5 #include<string.h>

 6 #include<stdlib.h>

 7 #include<algorithm>

 8 #include<iostream>

 9 #include<queue>

10 //#include<map>

11 #include<math.h>

12 using namespace std;

13 typedef long long ll;

14 //typedef __int64 int64;

15 const int maxn = 305;

16 const int inf = 0x7fffffff;

17 const double pi=acos(-1.0);

18 int a[ maxn ],b[ maxn ];

19 int vis_b[ maxn ];

20 struct node{

21     int x,y;

22 };

23 queue<node>step[ maxn ];

24 void init(){

25     for( int i=0;i<maxn;i++ ){

26         while( !step[ i ].empty() )

27             step[ i ].pop();

28     }

29     memset( vis_b,0,sizeof( vis_b ) );

30 }

31 int main(){

32     int n;

33     while( scanf("%d",&n)!=EOF ){

34         for( int i=0;i<n;i++ )

35             scanf("%d",&a[ i ]);

36         for( int i=0;i<n;i++ )

37             scanf("%d",&b[ i ]);

38         init();

39         int cnt=0;

40         node now;

41         for( int i=0;i<n;i++ ){

42             for( int j=0;j<n;j++ ){

43                 if( vis_b[ j ]==0&&a[ i ]==b[ j ] ){

44                     vis_b[ i ]=1;

45                     if( i==j ){

46                         //vis_b[ j ]=1;

47                         break;

48                     }

49                     else if( i<j ){

50                         //vis_b[ j ]=1;

51                         now.x=j-1,now.y=j;

52                         step[ cnt ].push( now );

53                         swap( b[ now.x ],b[ now.y ] );

54                         while( 1 ){

55                             if( now.x==i ) break;

56                             now.x--;

57                             now.y--;

58                             step[ cnt ].push( now );

59                             swap( b[ now.x ],b[ now.y ] );

60                         }

61                         cnt++;

62                         break;

63                     }

64                     else if( i>j ){

65                         //vis_b[ j ]=1;

66                         now.x=j,now.y=j+1;

67                         step[ cnt ].push( now );

68                         swap( b[ now.x ],b[ now.y ] );

69                         while( 1 ){

70                             if( now.y==i ) break;

71                             now.x++;

72                             now.y++;

73                             step[ cnt ].push( now );

74                             swap( b[ now.x ],b[ now.y ] );

75                         }

76                         cnt++;

77                         break;

78                     }

79                 }

80             }

81         }

82         int ans=0;

83         for( int i=0;i<cnt;i++ )

84             ans+=( step[ i ].size() );

85         printf("%d\n",ans);

86         for( int i=0;i<cnt;i++ ){

87             while( !step[ i ].empty() ){

88                 now=step[ i ].front();

89                 printf("%d %d\n",1+now.x,1+now.y);

90                 step[ i ].pop();

91             } 

92         }

93     }

94     return 0;

95 }