Shaolin

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 3021    Accepted Submission(s): 1273


Problem Description
Shaolin temple is very famous for its Kongfu monks.A lot of young men go to Shaolin temple every year, trying to be a monk there. The master of Shaolin evaluates a young man mainly by his talent on understanding the Buddism scripture, but fighting skill is also taken into account.
When a young man passes all the tests and is declared a new monk of Shaolin, there will be a fight , as a part of the welcome party. Every monk has an unique id and a unique fighting grade, which are all integers. The new monk must fight with a old monk whose fighting grade is closest to his fighting grade. If there are two old monks satisfying that condition, the new monk will take the one whose fighting grade is less than his.
The master is the first monk in Shaolin, his id is 1，and his fighting grade is 1,000,000,000.He just lost the fighting records. But he still remembers who joined Shaolin earlier, who joined later. Please recover the fighting records for him.
 

 

Input
There are several test cases.
In each test case:
The first line is a integer n (0 <n <=100,000),meaning the number of monks who joined Shaolin after the master did.(The master is not included).Then n lines follow. Each line has two integer k and g, meaning a monk's id and his fighting grade.( 0<= k ,g<=5,000,000)
The monks are listed by ascending order of jointing time.In other words, monks who joined Shaolin earlier come first.
The input ends with n = 0.
 

 

Output
A fight can be described as two ids of the monks who make that fight. For each test case, output all fights by the ascending order of happening time. Each fight in a line. For each fight, print the new monk's id first ,then the old monk's id.
 

 

Sample Input
3 2 1 3 3 4 2 0
 

 

Sample Output
2 1 3 2 4 2
 

 

Source
2013ACM-ICPC杭州赛区全国邀请赛 
题意：求前驱后继板子题
题解：ｏｒｚ
  1 #include <bits/stdc++.h>
  2 #define ll  __int64
  3 using namespace  std;
  4 ll tim=0,n,root;
  5 bool flag;
  6 struct node
  7 {
  8     ll father,left,right,data;
  9     ll id;
 10 } tree[100005];
 11  ll mins(ll aaa, ll bbb)
 12 {
 13     if(aaa<bbb)
 14         return aaa;
 15     else
 16         return bbb;
 17 }
 18  ll abss(ll x)
 19 {
 20     if(x<0)
 21         return -x;
 22     else
 23         return x;
 24 }
 25 void rightrotate(ll x)
 26 {
 27     ll y=tree[x].father;
 28     ll z=tree[y].father;
 29     tree[y].left=tree[x].right;
 30     if(tree[x].right!=-1)
 31     {
 32         tree[tree[x].right].father=y;
 33     }
 34     tree[x].father=z;
 35     if(z!=-1)
 36     {
 37         if(tree[z].left==y) tree[z].left=x;
 38         else tree[z].right=x;
 39     }
 40     tree[x].right=y;
 41     tree[y].father=x;
 42 }
 43 void leftrotate(ll x)
 44 {
 45     ll y=tree[x].father;
 46     ll z=tree[y].father;
 47     tree[y].right=tree[x].left;
 48     if(tree[x].left!=-1)
 49     {
 50         tree[tree[x].left].father=y;
 51     }
 52     tree[x].father=z;
 53     if(z!=-1)
 54     {
 55         if(tree[z].left==y) tree[z].left=x;
 56         else tree[z].right=x;
 57     }
 58     tree[x].left=y;
 59     tree[y].father=x;
 60 }
 61 void splay(ll x)
 62 {
 63     while(tree[x].father!=-1)
 64     {
 65         ll y=tree[x].father;
 66         ll z=tree[y].father;
 67         if(z==-1)
 68         {
 69             if(tree[y].left==x) rightrotate(x);
 70             else leftrotate(x);
 71         }
 72         else
 73         {
 74             if(tree[z].left==y&&tree[y].left==x)
 75             {
 76                 rightrotate(y);
 77                 rightrotate(x);
 78             }
 79             else if(tree[z].left==y&&tree[y].right==x)
 80             {
 81                 leftrotate(x);
 82                 rightrotate(x);
 83             }
 84             else if(tree[z].right==y&&tree[y].right==x)
 85             {
 86                 leftrotate(y);
 87                 leftrotate(x);
 88             }
 89             else
 90             {
 91                 rightrotate(x);
 92                 leftrotate(x);
 93             }
 94         }
 95     }root=x;
 96 }
 97 ll qq(ll x)
 98 {
 99     ll y=tree[x].left;
100     if(y==-1) return y;
101     while(tree[y].right!=-1) {
102             y=tree[y].right;
103     }
104     return y;
105 }
106 ll hj(ll x)
107 {
108     ll y=tree[x].right;
109     if(y==-1) return y;
110     while(tree[y].left!=-1){
111             y=tree[y].left;
112     }
113     return y;
114 }
115 int BST_insert(ll idd,ll dat,ll x)
116 {
117     if(dat==tree[x].data)
118     {
119         flag=false ;
120         splay(x);
121         return 0;
122     }
123     if(dat<tree[x].data)
124     {
125         if(tree[x].left==-1)
126         {
127             tree[x].left=tim;
128             tree[tim].father=x;
129             tree[tim].left=tree[tim].right=-1;
130             tree[tim].data=dat;
131             tree[tim].id=idd;
132         }
133         else
134             BST_insert(idd,dat,tree[x].left);
135     }
136     else
137     {
138         if(tree[x].right==-1)
139         {
140             tree[x].right=tim;
141             tree[tim].father=x;
142             tree[tim].left=tree[tim].right=-1;
143             tree[tim].data=dat;
144             tree[tim].id=idd;
145         }
146         else
147             BST_insert(idd,dat,tree[x].right);
148     }
149 }
150 ll insert1(ll idd,ll dat)
151 {
152     flag=true;
153     tim++;
154     BST_insert(idd,dat,root);
155     if(flag==false) return 0;
156     splay(tim);
157     ll q=qq(tim);
158     ll h=hj(tim);
159     ll minx=2000000000;
160     ll iddd=0;
161     if(q!=-1) {
162        if(minx>abss(tree[q].data-dat)){
163            minx=abss(tree[q].data-dat);
164            iddd=tree[q].id;
165        }
166     }
167     if(h!=-1) {
168         if(minx>abss(tree[h].data-dat)){
169            minx=abss(tree[h].data-dat);
170            iddd=tree[h].id;
171        }
172     }
173     printf("%I64d %I64d\n",idd,iddd);
174 }
175 int main()
176 {
177     int n;
178     ll aa=0;
179     while(scanf("%d",&n)!=EOF)
180     {
181     if(n==0)
182         return 0;
183     tim=0;
184     tim++;
185     tree[tim].father=-1;
186     tree[tim].left=tree[tim].right=-1;
187     tree[tim].data=1000000000;
188     tree[tim].id=1;
189     root=tim;
190     for(ll i=1; i<=n; i++)
191     {
192         ll aa=0,bb;
193         scanf("%I64d %I64d",&aa,&bb);
194         insert1(aa,bb);
195     }
196     }
197     return 0;
198 }