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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 dont 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 multiset

 ```1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 ``` ```#include #include #include #include #include #include #include   using namespace std;   const int N=100005; struct node { int l,r; int sum; }mem[N*3];   int num[N];// int place,suml,sumr;// int to;// 1 0 int ans;// int L,R;// int Search(int,int ); void insert(int ,int ,int ); void Right()// { to=1; sumr-=num[R]; insert(1,R,-num[R]); --num[R]; ans=ans+R-place; place=R; } void Left()// { 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)// { 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)//p k k { 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)//d { 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)// insert(1,x,1); if(xplace) ++sumr; }else { if(num[place]>0)// { --num[place]; continue; } if(suml==0&&sumr==0)// continue; if(sumr==0)// { L=Search(1,suml); Left(); }else if(suml==0)// { R=Search(1,1); Right(); }else if(suml>0&&sumr>0) { L=Search(1,suml);// R=Search(1,suml+1);// if(place-L
 ```1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 ``` ```#include #include #include using namespace std; priority_queue,greater > minQue; priority_queue 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-leftright-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<

multiset:

 ```1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 ``` ```#include #include #include #include #include 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; multisets; //multisetset multiset::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)0){ Min=abs((*It)-cur); p=(*It); del=It; } } } s.erase(del); sum+=Min; pre=cur; // cur=p; // } } printf("Case %d: %I64d\n",Case,sum); } // system("pause"); return 0; }```