Shaolin
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 3191 Accepted Submission(s): 1350
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.
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.
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
Sample Output
2 1
3 2
4 2
题目链接：HDU 4585
题意就是少林寺里有一个初始的人，他的id是1，武力值是1e9，然后按照输入顺序有n个人来少林寺挑战，输出n行，对应每一个来挑战的人给他分配的对手是谁。如何分配对手的规则题目已经讲了。
简单的做法就是建一颗Treap，然后在树上进行二分答案，找前驱与后继再比较判断得出答案即可，水题一道
代码：
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
#define fin(name) freopen(name,"r",stdin)
#define fout(name) freopen(name,"w",stdout)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
typedef pair<int, int> pii;
typedef long long LL;
const double PI = acos(-1.0);
const int N = 100010;
struct Treap
{
int ls, rs, v, sz, id;
int rnd;
};
Treap T[N];
int rt, tot; void init()
{
rt = 0;
tot = 0;
}
void pushup(int k)
{
T[k].sz = T[T[k].ls].sz + T[T[k].rs].sz + 1;
}
void lturn(int &k)
{
int rs = T[k].rs;
T[k].rs = T[rs].ls;
T[rs].ls = k;
T[rs].sz = T[k].sz;
pushup(k);
k = rs;
}
void rturn(int &k)
{
int ls = T[k].ls;
T[k].ls = T[ls].rs;
T[ls].rs = k;
T[ls].sz = T[k].sz;
pushup(k);
k = ls;
}
void ins(int &k, int v, int id)
{
if (!k)
{
k = ++tot;
T[k].ls = T[k].rs = 0;
T[k].v = v, T[k].id = id;
T[k].sz = 1;
T[k].rnd = rand();
}
else
{
++T[k].sz;
if (v < T[k].v)
{
ins(T[k].ls, v, id);
if (T[T[k].ls].rnd < T[k].rnd)
rturn(k);
}
else
{
ins(T[k].rs, v, id);
if (T[T[k].rs].rnd < T[k].rnd)
lturn(k);
}
}
}
void get_pre(int k, int v, int &ans)
{
if (!k)
return ;
if (T[k].v < v)
{
ans = k;
get_pre(T[k].rs, v, ans);
}
else
get_pre(T[k].ls, v, ans);
}
void get_post(int k, int v, int &ans)
{
if (!k)
return ;
if (T[k].v > v)
{
ans = k;
get_post(T[k].ls, v, ans);
}
else
get_post(T[k].rs, v, ans);
}
int main(void)
{
srand(987654321);
int n;
while (~scanf("%d", &n) && n)
{
init();
ins(rt, 1e9, 1);
while (n--)
{
int id, v;
scanf("%d%d", &id, &v);
int pre = -1, post = -1;
get_pre(rt, v, pre);
get_post(rt, v, post);
if (pre == -1)
printf("%d %d\n", id, T[post].id);
else if (post == -1)
printf("%d %d\n", id, T[pre].id);
else
{
int dxpre = abs(v - T[pre].v), dxpost = abs(v - T[post].v);
if (dxpre < dxpost)
printf("%d %d\n", id, T[pre].id);
else if (dxpre > dxpost)
printf("%d %d\n", id, T[post].id);
else
printf("%d %d\n", id, T[pre].id);
}
ins(rt, v, id);
}
}
return 0;
}