hdu------(4302）Holedox Eating(樹狀數組+二分)

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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 DescriptionHoledox 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. 

 

InputThe 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. 

 

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

 

Sample Input310 80 10 510 20 0111 10 70 10 510 20 01110 80 10 10 510 20 011 

 

Sample OutputCase 1: 9Case 2: 4Case 3: 2 

 

AuthorBUPT 

 

Source 2012 Multi-University Training Contest 1    代碼：

  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; //開始從0開始所以必須去右邊 35     pos=1;  //初始牛的位置在1初開始數 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說明那條牛要吃蛋糕了 45          //但是不知道那邊有蛋糕... 46          int left=pos, right=len+1; 47          int mid; 48          int rr=inf;   //記錄最近的右邊蛋糕位置 49          int ll=-inf;   //記錄最近的左邊蛋糕位置 50          bool iseat_r=false; 51          int st_num=query(pos-1);  //開始位置以下的位置蛋糕的數目 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)  //又一邊有蛋糕吃就可以了，否則什麼都不做 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，真的可以隨便吃嗎,以繼續保持原來的方向優先 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);  //吃掉了一個蛋糕 99        }100      }101    }102    printf("Case %d: %d\n",i,ans);103   }104 return 0;105 }

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hdu------(4302）Holedox Eating(樹狀數組+二分)