POJ2201+RMQ

 1 /*

 2 RMQ

 3 */

 4 #include<stdio.h>

 5 #include<string.h>

 6 #include<stdlib.h>

 7 #include<algorithm>

 8 #include<iostream>

 9 #include<math.h>

10 using namespace std;

11 const int maxn = 50005;

12 struct Node{

13     int ki,ai,id;

14 }node[ maxn ];

15 int dp[ maxn ][ 24 ];

16 int bit[ 32 ];

17 int ans[ maxn ][ 3 ];

18 

19 void init(){

20     memset( ans,0,sizeof( ans ) );

21     bit[0] = 1;

22     for( int i=1;i<32;i++ )

23         bit[ i ] = bit[ i-1 ]*2;

24     return ;

25 }

26 

27 int cmp( Node a,Node b ){

28     return a.ki<b.ki;

29 }

30 

31 void ST( int n ){

32     int K = (int)(log( 1.0*n )/log( 2.0 ));

33     for( int i=1;i<=n;i++ )

34         dp[ i ][ 0 ] = i;//存储着i，j之间ai值最小的位置

35     for( int j=1;j<=K;j++ ){

36         for( int i=1;i<=n-bit[j]+1;i++ ){

37             if( node[ dp[i][j-1] ].ai<node[ dp[i+bit[j-1]][j-1] ].ai )

38                 dp[ i ][ j ] = dp[i][j-1];

39             else

40                 dp[ i ][ j ] = dp[i+bit[j-1]][j-1];

41         }

42     }

43 }

44 

45 int RMQ( int l,int r ){

46     int k = (int)(log((1+r-l)*1.0)/log(2.0));

47     if( node[ dp[l][k] ].ai<node[ dp[r-bit[k]+1][k] ].ai ) return dp[l][k];

48     else return dp[r-bit[k]+1][k];

49 }

50 

51 int solve( int fa,int l,int r ){

52     if( l>r ) return 0;

53     int pos = RMQ( l,r );

54     int id = node[pos].id;

55     ans[ id ][ 0 ] = fa;

56     ans[ id ][ 1 ] = solve( id,l,pos-1 );

57     ans[ id ][ 2 ] = solve( id,pos+1,r );

58     return id;

59 }

60 

61 int main(){

62     int n;

63     init();

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

65         for( int i=1;i<=n;i++ ){

66             scanf("%d%d",&node[i].ki,&node[i].ai);

67             node[i].id = i;

68         }

69         sort( node+1,node+n+1,cmp );

70         ST( n );

71         solve( 0,1,n );

72         puts("YES");

73         for( int i=1;i<=n;i++ ){

74             printf("%d %d %d\n",ans[i][0],ans[i][1],ans[i][2]);

75         }

76     }

77     return 0;

78 }

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改变了dp中存储的内容

 1 /*

 2 RMQ

 3 */

 4 #include<stdio.h>

 5 #include<string.h>

 6 #include<stdlib.h>

 7 #include<algorithm>

 8 #include<iostream>

 9 #include<math.h>

10 using namespace std;

11 const int maxn = 50005;

12 struct Node{

13     int ki,ai,id;

14 }node[ maxn ];

15 int dp[ maxn ][ 24 ];

16 int bit[ 32 ];

17 int ans[ maxn ][ 3 ];

18 int Ai[ 70001 ];

19 const int MAX = 30000;

20 

21 void init(){

22     memset( ans,0,sizeof( ans ) );

23     bit[0] = 1;

24     for( int i=1;i<32;i++ )

25         bit[ i ] = bit[ i-1 ]*2;

26     return ;

27 }

28 

29 int cmp( Node a,Node b ){

30     return a.ki<b.ki;

31 }

32 

33 void ST( int n ){

34     int K = (int)(log( 1.0*n )/log( 2.0 ));

35     for( int i=1;i<=n;i++ )

36         dp[ i ][ 0 ] = node[i].ai;//存储着i，j之间ai值最小的位置

37     for( int j=1;j<=K;j++ ){

38         for( int i=1;i<=n-bit[j]+1;i++ ){

39             dp[ i ][ j ] = min( dp[i][j-1],dp[i+bit[j-1]][j-1]);

40         }

41     }

42 }

43 

44 int RMQ( int l,int r ){

45     int k = (int)(log((1+r-l)*1.0)/log(2.0));

46     int value = min( dp[l][k],dp[r-bit[k]+1][k] );

47     return Ai[ value+MAX ];

48 }

49 

50 int solve( int fa,int l,int r ){

51     if( l>r ) return 0;

52     int pos = RMQ( l,r );

53     int id = node[pos].id;

54     ans[ id ][ 0 ] = fa;

55     ans[ id ][ 1 ] = solve( id,l,pos-1 );

56     ans[ id ][ 2 ] = solve( id,pos+1,r );

57     return id;

58 }

59 

60 int main(){

61     int n;

62     init();

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

64         for( int i=1;i<=n;i++ ){

65             scanf("%d%d",&node[i].ki,&node[i].ai);

66             node[i].id = i;

67             //Ai[ node[i].ai ] = i;

68         }

69         sort( node+1,node+n+1,cmp );

70         for( int i=1;i<=n;i++ )

71             Ai[ node[i].ai+MAX ] = i;

72         ST( n );

73         solve( 0,1,n );

74         puts("YES");

75         for( int i=1;i<=n;i++ ){

76             printf("%d %d %d\n",ans[i][0],ans[i][1],ans[i][2]);

77         }

78     }

79     return 0;

80 }

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