題目描述

You have an array: a1, a2, , an and you must answer for some queries.
For each query, you are given an interval [L, R] and two numbers p and K. Your goal is to find the Kth closest distance between p and aL, aL+1, ..., aR.
The distance between p and ai is equal to |p - ai|.
For example:
A = {31, 2, 5, 45, 4 } and L = 2, R = 5, p = 3, K = 2.
|p - a2| = 1, |p - a3| = 2, |p - a4| = 42, |p - a5| = 1.
Sorted distance is {1, 1, 2, 42}. Thus, the 2nd closest distance is 1.

輸入

The first line of the input contains an integer T (1 <= T <= 3) denoting the number of test cases.
For each test case:
The first line contains two integers n and m (1 <= n, m <= 10^5) denoting the size of array and number of queries.
The second line contains n space-separated integers a1, a2, ..., an (1 <= ai <= 10^6). Each value of array is unique.
Each of the next m lines contains four integers L', R', p' and K'.
From these 4 numbers, you must get a real query L, R, p, K like this:?
L = L' xor X, R = R' xor X, p = p' xor X, K = K' xor X, where X is just previous answer and at the beginning, X = 0.
(1 <= L < R <= n, 1 <= p <= 10^6, 1 <= K <= 169, R - L + 1 >= K).

輸出

For each query print a single line containing the Kth closest distance between p and aL, aL+1, ..., aR.

樣例輸入

1 5 2 31 2 5 45 4 1 5 5 1 2 5 3 2

樣例輸出

0 1
題意 給一個序列A，每次詢問L，R，p，k，輸出[L,R]區間中第k大的|p-a[i]| 思路 二分答案，查詢[L,R]中[p-mid,p+mid]的數的個數 可以用主席樹/歸并樹維護區間x，y之間數的個數（題解說還可以用線段樹） 主席樹O(mlog1e6logn) 歸并樹O(mlog1e6logn^2)
歸并樹：https://www.cnblogs.com/bennettz/p/8342242.html #include <bits/stdc++.h> #define ll long long using namespace std; const int N=1e5+10; int T,n,m; int a[N],t[20][N]; int read() {int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f; } void build(int s,int l,int r) {if (l==r){t[s][l]=a[l];return;}int mid=(l+r)>>1;build(s+1,l,mid); build(s+1,mid+1,r);for (int i=l,j=mid+1,k=l;i<=mid||j<=r;){if (j>r) t[s][k++]=t[s+1][i++];else if (i>mid || t[s+1][i]>t[s+1][j]) t[s][k++]=t[s+1][j++];else t[s][k++]=t[s+1][i++];} } int query(int s,int l,int r,int L,int R,int x,int y) {if (x>y ) return 0;if (L<=l&r<=R){//printf("x=%d,y=%d,l=%d,r=%d,>y=%d,>=x=%d\n",x,y,l,r,upper_bound(t[s]+l,t[s]+r+1,y),upper_bound(t[s]+l,t[s]+r+1,x));return upper_bound(t[s]+l,t[s]+r+1,y)-lower_bound(t[s]+l,t[s]+r+1,x);}int mid=(l+r)>>1,ans=0;if (L<=mid) ans+=query(s+1,l,mid,L,R,x,y);if (R>mid) ans+=query(s+1,mid+1,r,L,R,x,y);return ans; } int main() {// freopen("14162.in","r",stdin);// freopen("1.out","w",stdout);T=read();while(T--){n=read(); m=read();//memset(t,0,sizeof(t));for (int i=1;i<=n;i++) a[i]=read();build(0,1,n);int ans=0;int L,R,p,k;while (m--){L=read(); R=read(); p=read(); k=read();L^=ans; R^=ans; p^=ans; k^=ans;int l=0,r=1000005;while (l<=r){int mid=(l+r)>>1;//cout<<mid<<endl;if (query(0,1,n,L,R,p-mid,p+mid)>=k) ans=mid,r=mid-1;else l=mid+1;}printf("%d\n",ans);}}// fclose(stdin);// fclose(stdout);return 0; } 歸并樹 #include <bits/stdc++.h> using namespace std; const int N=1e5+10; int ls[N*21],rs[N*21],s[N*21],root[N]; int a[N],b[N]; int T,n,m,sz; int read() {int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f; } void Insert(int l,int r,int x,int &y,int v) {y=++sz;s[y]=s[x]+1;if (l==r) return;ls[y]=ls[x]; rs[y]=rs[x];int mid=(l+r)>>1;if (v<=mid) Insert(l,mid,ls[x],ls[y],v);else Insert(mid+1,r,rs[x],rs[y],v); } int query(int l,int r,int L,int R,int x,int y) {if (L<=l&&r<=R) return s[y]-s[x];int ret=0;int mid=(l+r)>>1;if (L<=mid) ret+=query(l,mid,L,R,ls[x],ls[y]);if (R>mid) ret+=query(mid+1,r,L,R,rs[x],rs[y]);return ret; }int main() {T=read();while(T--){sz=0;n=read(); m=read();for (int i=1;i<=n;i++) a[i]=read(),b[i]=a[i];sort(b+1,b+1+n);int cnt=unique(b+1,b+1+n)-b-1;for (int i=1;i<=n;i++){a[i]=lower_bound(b+1,b+1+cnt,a[i])-b+1;Insert(1,N,root[i-1],root[i],a[i]);}int ans=0;int L,R,p,k;while (m--){L=read(); R=read(); p=read(); k=read();L^=ans,R^=ans,p^=ans,k^=ans;int l=0,r=1000006;while (l<=r){int mid=(l+r)>>1;int ll=lower_bound(b+1,b+1+cnt,p-mid)-b+1;int rr=upper_bound(b+1,b+1+cnt,p+mid)-b;if (query(1,N,ll,rr,root[L-1],root[R])>=k) ans=mid,r=mid-1;else l=mid+1;}printf("%d\n",ans);}}return 0; } 主席樹

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轉載于:https://www.cnblogs.com/tetew/p/11286580.html

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