hdu------(4302）Holedox Eating(树状数组+二分)-优快云博客

本文探讨了一种关于小型动物Holedox在管道中寻找并进食蛋糕的路径优化算法，详细介绍了输入输出规则，算法流程及求解过程，旨在最小化Holedox的总移动距离。

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Holedox Eating

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3362 Accepted Submission(s): 1145

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
  

Author

BUPT

Source

2012 Multi-University Training Contest 1

代码：

  1 #include<cstring>
  2 #include<cstdio>
  3 #define maxn 100080
  4 #define inf 0x3f3f3f3f
  5 int next[maxn];
  6 int len,n;
  7 int lowbit(int x){
  8    return x&(-x);
  9 }
 10 void update(int st,int val){
 11   while(st<=len+1){
 12       next[st]+=val;
 13     st+=lowbit(st);
 14   }
 15 }
 16 int query(int st){
 17     int ans=0;
 18     while(st>0){
 19       ans+=next[st];
 20        st-=lowbit(st);
 21     }
 22   return ans;
 23 }
 24 
 25 int main(){
 26   int test;
 27   int jud,tem,pos,ans;
 28   //freopen("test.in","r",stdin);
 29   //freopen("test1.in","w",stdout);s
 30   scanf("%d",&test);
 31  for(int i=1;i<=test;i++ ){
 32       scanf("%d%d",&len,&n);
 33     memset(next,0,sizeof(next));
 34        bool flag=true; //开始从0开始所以必须去右边
 35     pos=1;  //初始牛的位置在1初开始数
 36     ans=0;
 37   while(n--)
 38   {
 39      scanf("%d",&jud);
 40      if(!jud){
 41        scanf("%d",&tem);
 42        update(tem+1,1);
 43      }
 44      else{  //如果jud=1说明那条牛要吃蛋糕了
 45          //但是不知道那边有蛋糕...
 46          int left=pos, right=len+1;
 47          int mid;
 48          int rr=inf;   //记录最近的右边蛋糕位置
 49          int ll=-inf;   //记录最近的左边蛋糕位置
 50          bool iseat_r=false;
 51          int st_num=query(pos-1);  //开始位置以下的位置蛋糕的数目
 52          while(left<=right){
 53            mid=left+((right-left)>>1);
 54            if(query(mid)>st_num) {
 55              right=mid-1;
 56              rr=mid;
 57              iseat_r=true;  //表示吃到了蛋糕
 58            }
 59            else
 60              left=mid+1;
 61          }
 62          left=1;
 63          right=pos;
 64          st_num=query(pos);
 65          bool iseat_l=false;
 66         while(left<=right){
 67           mid=left+((right-left)>>1);
 68           if(st_num-query(mid-1)>0){
 69               left=mid+1;
 70             ll=mid;
 71             iseat_l=true;
 72           }
 73           else  right=mid-1;
 74         }
 75       if(iseat_l||iseat_r)  //又一边有蛋糕吃就可以了，否则什么都不做
 76       {
 77         int len1=pos-ll;
 78         int len2=rr-pos;
 79         if(len1<len2){
 80             pos=ll;
 81             ans+=len1;
 82             flag=false;
 83         }
 84         else { //如果相等，真的可以随便吃吗,以继续保持原来的方向优先
 85             if(len1>len2){
 86                 pos=rr;
 87                 ans+=len2;
 88                 flag=true;
 89             }
 90             else if(flag){ pos=rr;
 91                 ans+=len2;
 92             }
 93          else{
 94             pos=ll;
 95             ans+=len1;
 96             }
 97         }
 98            update(pos,-1);  //吃掉了一个蛋糕
 99        }
100      }
101    }
102    printf("Case %d: %d\n",i,ans);
103   }
104 return 0;
105 }