hdu4585 Treap与名次树/STL map(C/C++)

hdu4585

题目地址：https://acm.dingbacode.com/showproblem.php?pid=4585

Shaolin

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)

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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.

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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.

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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.

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Sample Input

3
2 1
3 3
4 2
0

Sample Output

2 1
3 2
4 2

题意

先对老和尚的等级排序，在加入一个新和尚是，找到等级最接近的老和尚，输出老和尚的id。

总复杂度O(nlong n)

<解法一> STL map

AC代码

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int maxn=1e9;
 4 map<int,int> mp;//it->first是等级，it->second是id
 5 int main(){
 6     int n;
 7     while(~scanf("%d",&n)&&n){
 8         mp.clear();
 9         mp[maxn]=1;//方丈1，等级是1 000 000 000
10         while(n--){
11             int id,g;
12             scanf("%d%d",&id,&g);//新和尚id，等级是g
13             mp[g]=id;//进队
14             int ans;
15             map<int,int>::iterator it=mp.find(g);//找到排好序的位置
16             if(it==mp.begin()) ans=(++it)->second;
17             else{
18                 map<int,int>::iterator it2=it;
19                 it2--;it++;//等级接近的前后两个老和尚
20                 if(g-it2->first<=it->first-g)
21                     ans=it2->second;
22                 else ans=it->second;
23             }
24             printf("%d %d\n",id,ans);
25         }
26     }
27     return 0;
28 }

<解法二> Treap树

AC代码

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int id[5000005];
 4 struct node{
 5     int size;//以这个结点为根的子树的结点总数量，用于名次树
 6     int rank;//优先级
 7     int key;//键值
 8     node *son[2];//son[0]是左儿子，son[1]是右儿子
 9     bool operator<(const node &a)const{return rank<a.rank;}
10     int cmp(int x)const{
11         if(x==key) return -1;
12         return x<key?0:1;
13     }
14     void update(){//更新size
15         size=1;
16         if(son[0]!=NULL) size+=son[0]->size;
17         if(son[1]!=NULL) size+=son[1]->size;
18     }
19 };
20 void rotate(node *&o,int d){//d=0，左旋;d=1，右旋
21     node *k=o->son[d^1];//d^1与1-d等价，但是更快
22     o->son[d^1]=k->son[d];
23     k->son[d]=o;
24     o->update();
25     k->update();
26     o=k;
27 }
28 void insert(node *&o,int x){//把x插入到树中
29     if(o==NULL){
30         o=new node();
31         o->son[0]=o->son[1]=NULL;
32         o->rank=rand();
33         o->key=x;
34         o->size=1;
35     }
36     else{
37         int d=o->cmp(x);
38         insert(o->son[d],x);
39         o->update();
40         if(o<o->son[d]) rotate(o,d^1);
41     }
42 }
43 int kth(node *o,int k){//返回第k大的数
44     if(o==NULL||k<=0||k>o->size) return -1;
45     int s=o->son[1]==NULL?0:o->son[1]->size;
46     if(k==s+1) return o->key;
47     else if(k<=s) return kth(o->son[1],k);
48     else return kth(o->son[0],k-s-1);
49 }
50 int find(node *o,int k){//返回元素k的名次
51     if(o==NULL) return -1;
52     int d=o->cmp(k);
53     if(d==-1) return o->son[1]==NULL?1:o->son[1]->size+1;
54     else if(d==1) return find(o->son[d],k);
55     else{
56         int tmp=find(o->son[d],k);
57         if(tmp==-1) return -1;
58         else return o->son[1]==NULL?tmp+1:tmp+1+o->son[1]->size;
59     }
60 }
61 int main(){
62     int n;
63     while(~scanf("%d",&n)&&n){
64         srand(time(NULL));
65         int k,g;
66         scanf("%d%d",&k,&g);
67         node *root=new node();
68         root->son[0]=root->son[1]=NULL;
69         root->rank=rand();
70         root->key=g;
71         root->size=1;
72         id[g]=k;
73         printf("%d 1\n",k);
74         for(int i=2;i<=n;i++){
75             scanf("%d%d",&k,&g);
76             id[g]=k;
77             insert(root,g);
78             int t=find(root,g);//返回新和尚的名次
79             int ans1,ans2,ans;
80             ans1=kth(root,t-1);//前一名的老和尚
81             ans2=kth(root,t+1);//后一名的老和尚
82             if(ans1!=-1&&ans2!=-1) ans=ans1-g>=g-ans2?ans2:ans1;
83             else if(ans1==-1) ans=ans2;
84             else ans=ans1;
85             printf("%d %d\n",k,id[ans]);
86         }
87     }
88     return 0;
89 }

感觉这段比较难理解的话，可以去看看《算法竞赛入门经典训练指南》（蓝书）P230的3.5.2

函数？拿来吧你

// 讲真，发生了一件比较诡异的事情，我在之前的博客里不是写过一次rotate函数嘛，然后把那边的rotate函数直接复制过来（当然要加上更新，而且必须是先维护o，再维护k）去交的话CE了，但是重打一遍就AC了。。。就离谱/笑哭//笑哭//笑哭/

// 上面提到的之前的博客的地址：https://www.cnblogs.com/ynzhang2020/p/15070994.html