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 }
View Code

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 }
View Code

 

posted @ 2013-07-14 12:18  xxx0624  阅读(248)  评论(0)  编辑  收藏  举报