傳送門 D. Sum of Medians time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output

In one well-known algorithm of finding the?k-th order statistics we should divide all elements into groups of five consecutive elements and find the median of each five. A median is called the middle element of a sorted array (it's the third largest element for a group of five). To increase the algorithm's performance speed on a modern video card, you should be able to find a sum of medians in each five of the array.

A?sum of medians?of a sorted?k-element set?S?=?{a1,?a2,?...,?ak}, where?a1?<?a2?<?a3?<?...?<?ak, will be understood by as

The??operator stands for taking the remainder, that is??stands for the remainder of dividing?x?by?y.

To organize exercise testing quickly calculating?the sum of medians?for a changing set was needed.

Input

The first line contains number?n?(1?≤?n?≤?105), the number of operations performed.

Then each of?n?lines contains the description of one of the three operations:

• add?x?— add the element?x?to the set;
• del?x?— delete the element?x?from the set;
• sum?— find the?sum of medians?of the set.

For any?add?x?operation it is true that the element?x?is not included in the set directly before the operation.

For any?del?x?operation it is true that the element?x?is included in the set directly before the operation.

All the numbers in the input are positive integers, not exceeding?109.

Output

For each operation?sum?print on the single line?the sum of medians?of the current set. If the set is empty, print 0.

Please, do not use the?%lld?specificator to read or write 64-bit integers in C++. It is preferred to use the?cin,?cout?streams (also you may use the?%I64d?specificator).

Sample test(s) input 6
add 4
add 5
add 1
add 2
add 3
sum output 3 input 14
add 1
add 7
add 2
add 5
sum
add 6
add 8
add 9
add 3
add 4
add 10
sum
del 1
sum output 5
11
13
----------------------------------------------------------------------------------------------- Solution 用線段樹維護集合 1.預處理所有add/del操作，將涉及的元素離散化為1...n 2.add x ，設x離散化為i，修改區間(i+1， n), 將各節點的sum[5]右移一位，再插入x 3.del x, 設x離散化為i, 修改區間(i+1, n), 將各節點的sum[5]左移一位, 在刪除x 4.插入/刪除元素需要知道元素在集合中的位置 (place),這由樹狀數組（BIT）維護。 5.對區間sum[5]的左移/右移操作需要用lazy-tag優化。 6.如果手寫離散化，要注意實現細節。 7.需考慮極端輸入，比如無sum詢問的輸入 --------------------------------------------------------------------------------------------- 這是我第一發提交，TLE on test 5, 跪在極端輸入上了，TLE的原因是調用了add(0, 1) #include<bits/stdc++.h> using namespace std; typedef long long ll; const int MAX_N=1e5+5; #define X first #define Y second //BIT int bit[MAX_N], rb; int sum(int i){int s=0;while(i){s+=bit[i];i-=i&-i;}return s; } void add(int i, int x){while(i<=rb){bit[i]+=x;i+=i&-i;} }typedef pair<int, int> P; P q[MAX_N]; int val[MAX_N]; char type[MAX_N][5]; int lisan(int N){int v;for(int i=0; i<N; i++){scanf("%s", type[i]);switch(*type[i]){case 'a':case 'd':scanf("%d", &v);q[i]=P(v, i);break; case 's':q[i]=P(0, i);　　//TLE在這里break;}}sort(q, q+N);int ord=0;for(int i=0; i<N; ord++){val[ord]=q[i].X;do{q[i].X=ord;i++;}while(i<N&&q[i].X==val[ord]);}return ord; } bool cmp(const P &a, const P &b){return a.Y<b.Y; } //ST struct node{int l, r, tag;ll sum[5];int mid(){return (l+r)>>1;} }T[MAX_N<<2]; void shift(int id, int cnt){cnt%=5;if(!cnt) return;ll tmp[5];memcpy(tmp, T[id].sum, sizeof(tmp));for(int i=0; i<5; i++){int nt=(i+cnt+5)%5;T[id].sum[nt]=tmp[i];} } void pushdown(int id){node &now=T[id], &lch=T[id<<1], &rch=T[id<<1|1];lch.tag+=now.tag, rch.tag+=now.tag;shift(id<<1, now.tag);shift(id<<1|1, now.tag);now.tag=0; //error prone } void unite(int id){node &now=T[id], &lch=T[id<<1], &rch=T[id<<1|1];for(int i=0; i<5; i++)now.sum[i]=lch.sum[i]+rch.sum[i]; } void build(int id, int l, int r){node &now=T[id];T[id].l=l, T[id].r=r, T[id].tag=0;memset(T[id].sum, 0, sizeof(T[id].sum));if(l==r) return;int mid=(l+r)>>1;build(id<<1, l, mid);build(id<<1|1, mid+1, r); } void insert(int id, int pos, int ord, int v){node &now=T[id];if(now.l==now.r){now.sum[ord%5]+=v;}else{pushdown(id);if(pos<=now.mid())insert(id<<1, pos, ord, v);elseinsert(id<<1|1, pos, ord, v);unite(id);} } void shift(int id, int lb, int cnt){if(lb>rb) return;node &now=T[id];if(now.l>=lb){shift(id, cnt);now.tag+=cnt; }else{pushdown(id);if(lb<=now.mid())shift(id<<1, lb, cnt);shift(id<<1|1, lb, cnt);unite(id);} }int main(){//freopen("in", "r", stdin);int N;scanf("%d", &N);rb=lisan(N);build(1, 1, rb);sort(q, q+N, cmp);for(int i=0; i<N; i++){if(*type[i]=='a'){int ord=sum(q[i].X)+1;add(q[i].X, 1);shift(1, q[i].X+1, 1);insert(1, q[i].X, ord, val[q[i].X]);}else if(*type[i]=='d'){int ord=sum(q[i].X);add(q[i].X, -1);shift(1, q[i].X+1, -1);insert(1, q[i].X, ord, -val[q[i].X]);}else{printf("%I64d\n", T[1].sum[3]);}}return 0; }

其實我已經考慮到這種無sum詢問的輸入，但沒注意到它會導致我的代碼TLE

AC version

#include<bits/stdc++.h> using namespace std; typedef long long ll; const int MAX_N=1e5+5; #define X first #define Y second //BIT int bit[MAX_N], rb; int sum(int i){int s=0;while(i){s+=bit[i];i-=i&-i;}return s; } void add(int i, int x){while(i<=rb){bit[i]+=x;i+=i&-i;} }typedef pair<int, int> P; P q[MAX_N]; int val[MAX_N]; char type[MAX_N][5]; int lisan(int N){int v;for(int i=0; i<N; i++){scanf("%s", type[i]);switch(*type[i]){case 'a':case 'd':scanf("%d", &v);q[i]=P(v, i);break; case 's':q[i]=P(INT_MAX, i);　　//注意這里的改動break;}}sort(q, q+N);int ord=0;for(int i=0; i<N;){++ord;val[ord]=q[i].X;do{q[i].X=ord;i++;}while(i<N&&q[i].X==val[ord]);}return ord; } bool cmp(const P &a, const P &b){return a.Y<b.Y; } //ST struct node{int l, r, tag;ll sum[5];int mid(){return (l+r)>>1;} }T[MAX_N<<2]; void shift(int id, int cnt){cnt%=5;if(!cnt) return;ll tmp[5];memcpy(tmp, T[id].sum, sizeof(tmp));for(int i=0; i<5; i++){int nt=(i+cnt+5)%5;T[id].sum[nt]=tmp[i];} } void pushdown(int id){node &now=T[id], &lch=T[id<<1], &rch=T[id<<1|1];lch.tag+=now.tag, rch.tag+=now.tag;shift(id<<1, now.tag);shift(id<<1|1, now.tag);now.tag=0; //error prone } void unite(int id){node &now=T[id], &lch=T[id<<1], &rch=T[id<<1|1];for(int i=0; i<5; i++)now.sum[i]=lch.sum[i]+rch.sum[i]; } void build(int id, int l, int r){node &now=T[id];T[id].l=l, T[id].r=r, T[id].tag=0;memset(T[id].sum, 0, sizeof(T[id].sum));if(l==r) return;int mid=(l+r)>>1;build(id<<1, l, mid);build(id<<1|1, mid+1, r); } void insert(int id, int pos, int ord, int v){node &now=T[id];if(now.l==now.r){now.sum[ord%5]+=v;}else{pushdown(id);if(pos<=now.mid())insert(id<<1, pos, ord, v);elseinsert(id<<1|1, pos, ord, v);unite(id);} } void shift(int id, int lb, int cnt){if(lb>rb) return;node &now=T[id];if(now.l>=lb){shift(id, cnt);now.tag+=cnt; }else{pushdown(id);if(lb<=now.mid())shift(id<<1, lb, cnt);shift(id<<1|1, lb, cnt);unite(id);} }int main(){//freopen("in", "r", stdin);int N;scanf("%d", &N);rb=lisan(N);build(1, 1, rb);sort(q, q+N, cmp);for(int i=0; i<N; i++){if(*type[i]=='a'){int ord=sum(q[i].X)+1;add(q[i].X, 1);shift(1, q[i].X+1, 1);insert(1, q[i].X, ord, val[q[i].X]);}else if(*type[i]=='d'){int ord=sum(q[i].X);add(q[i].X, -1);shift(1, q[i].X+1, -1);insert(1, q[i].X, ord, -val[q[i].X]);}else{printf("%I64d\n", T[1].sum[3]);}}return 0; }

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轉載于:https://www.cnblogs.com/Patt/p/4696384.html

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