线段树，堆，multiset：Holedox Eating

Holedox Eating

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1187    Accepted Submission(s): 391

Problem Description

Holedox is a small animal which can be considered as one point. It lives in a straight pipe whose length is L. Holedox can only move along the pipe. Cakes may appear anywhere in the pipe, from time to time. When Holedox wants to eat cakes, it always goes to the nearest one and eats it. If there are many pieces of cake in different directions Holedox can choose, Holedox will choose one in the direction which is the direction of its last movement. If there are no cakes present, Holedox just stays where it is.

 

Input

The input consists of several test cases. The first line of the input contains a single integer T (1 <= T <= 10), the number of test cases, followed by the input data for each test case.The first line of each case contains two integers L,n(1<=L,n<=100000), representing the length of the pipe, and the number of events.
The next n lines, each line describes an event. 0 x(0<=x<=L, x is a integer) represents a piece of cake appears in the x position; 1 represent Holedox wants to eat a cake.
In each case, Holedox always starts off at the position 0.

 

Output

Output the total distance Holedox will move. Holedox don’t need to return to the position 0.

 

Sample Input

3
10 8
0 1
0 5
1
0 2
0 0
1
1
1 

10 7
0 1
0 5
1
0 2
0 0
1
1

10 8
0 1
0 1
0 5
1
0 2
0 0
1
1

 

Sample Output

Case 1: 9
Case 2: 4
Case 3: 2

这道题神奇得可以用三种方法，线段树容易想到但难敲，堆相对容易，multiset这种奇淫巧计竟然也可以过：

线段树：

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<queue>
#include<cmath>
#include<stack>
#include<algorithm>

using namespace std;

const int N=100005;
struct node
{
    int l,r;
    int sum;
}mem[N*3];

int num[N];//记录某个位置的食物数量
int place,suml,sumr;//小动物的位置 和它左边 右边的食物数量
int to;//表示方向 1 向右 0 向左
int ans;//保存答案
int L,R;//搜索左边和右边最靠近小动物的食物位置
int Search(int,int );
void insert(int ,int ,int );
void Right()//选择右边的食物 进行的一些必要更新
{
    to=1;
    sumr-=num[R];
    insert(1,R,-num[R]);
    --num[R];
    ans=ans+R-place;
    place=R;
}
void Left()//选择左边的食物 进行的一些必要更新
{
    to=0;
    suml-=num[L];
    insert(1,L,-num[L]);
    --num[L];
    ans=ans+place-L;
    place=L;
}
void build(int x,int i,int j)//建树
{
    mem[x].l=i;
    mem[x].r=j;
    mem[x].sum=0;
    if(i==j)
    return ;
    int mid=(i+j)>>1;
    build(x*2,i,mid);
    build(x*2+1,mid+1,j);
}
void insert(int x,int p,int k)//在p这个位置 插入k个食物 k可以为负 用来减少操作
{
    int mid=(mem[x].l+mem[x].r)>>1;
    if(mem[x].l==mem[x].r)
    {
        mem[x].sum+=k;
        return ;
    }
    if(p<=mid)
    insert(x*2,p,k);
    else
    insert(x*2+1,p,k);
    mem[x].sum=mem[x*2].sum+mem[x*2+1].sum;
}
int Search(int x,int d)//搜索第d个食物的位置
{
    if(mem[x].l==mem[x].r)
    return mem[x].r;
    if(mem[x*2].sum>=d)
    return Search(x*2,d);
    else
    return Search(x*2+1,d-mem[x*2].sum);
}
int main()
{
   int T;
   scanf("%d",&T);
   for(int w=1;w<=T;++w)
   {
       int n,m;
       place=0,suml=0,sumr=0;
       to=1;
       ans=0;
       scanf("%d %d",&n,&m);
       build(1,0,n);
       memset(num,0,sizeof(num));
       while(m--)
       {
           int k,x;
           scanf("%d",&k);
           if(k==0)
           {
               scanf("%d",&x);
               ++num[x];
               if(x!=place)//插入位置 不是在小动物位置才更新线段树
               insert(1,x,1);
               if(x<place)
               ++suml;
               else if(x>place)
               ++sumr;
           }else
           {
               if(num[place]>0)//在小动物位置 直接减少 不需其他操作
               {
                   --num[place];
                   continue;
               }
               if(suml==0&&sumr==0)//没有食物
               continue;
               if(sumr==0)//右边没食物 选左边的
               {
                   L=Search(1,suml);
                   Left();
               }else
               if(suml==0)//选右边的
               {
                   R=Search(1,1);
                   Right();
               }else
               if(suml>0&&sumr>0)
               {
                   L=Search(1,suml);//求的左边最近食物位置
                   R=Search(1,suml+1);//求的右边最近食物位置
                   if(place-L<R-place||(place-L==R-place&&to==0))
                   Left();
                   else
                   Right();
               }
           }
       }
       printf("Case %d: %d\n",w,ans);
   }
   return 0;
}

堆：

#include<iostream>
#include<queue>
#include<algorithm>
using namespace std;
priority_queue<int,vector<int>,greater<int> > minQue;
priority_queue<int> maxQue;
int main(){
	long long res;
	int k,l,n,left,targ,val,right,curr,dir;
	while(scanf("%d",&k)!=EOF){
		for(int i=1;i<=k;i++){
			while(!minQue.empty()) minQue.pop();
			while(!maxQue.empty()) maxQue.pop();
			res=0;
			curr=0;
			dir=1;
			scanf("%d%d",&l,&n);
			while(n--){
				scanf("%d",&targ);
				if(targ){
					if(minQue.empty()){
						if(!maxQue.empty()){
							left=maxQue.top();
							maxQue.pop();
							res+=curr-left;
							if(curr!=left) dir=0;
							curr=left;
						}
					}
					else{
						if(maxQue.empty()){
							right=minQue.top();
							minQue.pop();
							res+=right-curr;
							curr=right;
							dir=1;
						}
						else{
							left=maxQue.top();
							right=minQue.top();
							if(curr-left<right-curr){
								res+=curr-left;
								if(curr!=left) dir=0;
								curr=left;
								maxQue.pop();
							}
							else if(curr-left>right-curr){
								res+=right-curr;
								dir=1;
								curr=right;
								minQue.pop();
							}
							else{
								if(dir){
									res+=right-curr;
									dir=1;
									curr=right;
									minQue.pop();
								}
								else{
									res+=curr-left;
									if(curr!=left) dir=0;
									curr=left;
									maxQue.pop();
								}
							}
						}
					}
				}
				else{
					scanf("%d",&val);
					if(val<=curr) maxQue.push(val);
					else minQue.push(val);
				}
			}
			printf("Case %d: ",i);
			cout<<res<<endl;
		}
	}
	return 0;
}

multiset:

#include<iostream>
#include<cstring>
#include<cstdio>
#include<set>
#include<cmath>
using namespace std;
int main(){
	int t;
	long long sum;
	int L,n;
	int a,b;
	long long Min,p;
	scanf("%d",&t);
	for(int Case=1;Case<=t;Case++){
		scanf("%d%d",&L,&n);
		sum=0;
		multiset<int>s;                       //multiset用红黑树来组织元素数据，题目中需要允许重复插入元素键值，故不用set
		multiset<int>::iterator It,del;
		int cur=0,pre=0;
		while(n--){
			 scanf("%d",&a);
			 if(a==0){
				 scanf("%d",&b);
				 s.insert(b);
			 }
			 else{
				  Min=0x3fffffff;
				  if(!s.size())
					  continue;
				  for(It=s.begin();It!=s.end();It++){
					  if(abs(*It-cur)<Min){   
						  Min=abs((*It)-cur);
						  p=(*It);           //得到距离最近的那个蛋糕的位置
						  del=It;            //标记要吃的蛋糕，以便于删除
					  }
					  else if(abs(*It-cur)==Min){   //当距离一样时取同方向的
						  if((*It-cur)*(cur-pre)>0){
							  Min=abs((*It)-cur);
							  p=(*It);
							  del=It;
						  }
					  }
				  }
				  s.erase(del);
				  sum+=Min;
				  pre=cur;                    //之前所在的位置
				  cur=p;                      //目前的位置
			 }
		}
		printf("Case %d: %I64d\n",Case,sum);
	}
//	system("pause");
	return 0;
}