K-th Closest Distance
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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6621
K-th Closest Distance
Time Limit: 20000/15000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 2426 Accepted Submission(s): 871
Problem Description
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.
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.
Input
The first line of the input contains an integer T (1 <= T <= 3) denoting the number of test cases.
For each test case:
冘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).
For each test case:
冘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).
Output
For each query print a single line containing the Kth closest distance between p and aL, aL+1, ..., aR.
Sample Input
1
5 2
31 2 5 45 4
1 5 5 1
2 5 3 2
Sample Output
0
1
Source
题目大意:T组样例,N个数 M次询问,L,R,P,K 问你L,R区间内 距离P第K近的距离是多少
思路:比赛的时候没啥思路,没写,看题解发现,主席树可以求区间内<=K的数有多少个 但是我们要求的是与某个值相减绝对值第K小啊,这有什么关系吗?
假设我们知道答案是几,那么此时求出来L,R区间内的数与P相减第K小是不是就是我们这个答案呢? 显然是的,但是问题是我们不知道答案啊
这就是二分的思想了,二分答案,查询范围在P-ans,P+ans之间的数有多少个,如果个数大于等于K的话,证明我们的ans还可以减小,一直这样下去
就会得到正确的答案了
看代码:
#include<iostream> #include<vector> #include<cstdio> #include<algorithm> #include<math.h> using namespace std; typedef long long LL; const int maxn=1e5+5; int N,M; int sz,cnt=0,sum=0; int root[maxn],a[maxn],v[maxn]; int L,R,P,K,X=0; struct Node int l,r,sum; Node() l=r=sum=0; T[maxn<<6]; void Init() cnt=0;root[0]=0;T[cnt].l=T[cnt].r=T[cnt].sum=0;X=0; void Read(int &n)//只能读入整数 n=0; int f=1;//用于记录正负 char ch=getchar(); while(ch<‘0‘||ch>‘9‘)// if(ch==‘-‘) f=-1; ch=getchar(); while(ch>=‘0‘&&ch<=‘9‘)//下面两种用法都可以 //n=(n<<1)+(n<<3)+(ch-‘0‘);// n=(n<<1)+(n<<3)+(ch^48);//相当于 x*10+ch-‘0‘; ch=getchar(); n=n*f;//切记乘以f (记录的是+ -) return ; void Update(int l,int r,int &x,int y,int pos) T[++cnt]=T[y];T[cnt].sum++;x=cnt; if(l==r) return ; int mid=(l+r)>>1; if(pos<=mid) Update(l,mid,T[x].l,T[y].l,pos); else Update(mid+1,r,T[x].r,T[y].r,pos); int getid(int x) return lower_bound(v+1,v+1+sz,x)-(v); void Query(int l,int r,int x,int y,int h) // cout<<"l:"<<l<<" r:"<<r<<" sum:"<<sum<<endl; if(l==r) if(h>=v[l]) sum+=T[x].sum-T[y].sum; return ; int mid=(l+r)>>1; if(h<=v[mid]) Query(l,mid,T[x].l,T[y].l,h); else sum+=T[T[x].l].sum-T[T[y].l].sum; Query(mid+1,r,T[x].r,T[y].r,h); bool judge(int x) sum=0;Query(1,sz,root[R],root[L-1],P+x); int sum1=sum; sum=0;Query(1,sz,root[R],root[L-1],P-x-1); int sum2=sum; return sum1-sum2>=K; int main() int T; // scanf("%d",&T); Read(T); int ca=1; while(T--) Init(); // scanf("%d%d",&N,&M); Read(N);Read(M); for(int i=1;i<=N;i++) scanf("%d",&a[i]);v[i]=a[i]; sort(v+1,v+1+N);sz=unique(v+1,v+1+N)-(v+1); // for(int i=1;i<=sz;i++) cout<<v[i]<<" ";cout<<endl; // for(int i=1;i<=N;i++) cout<<getid(a[i])<<" ";cout<<endl; for(int i=1;i<=N;i++) Update(1,sz,root[i],root[i-1],getid(a[i])); while(M--) sum=0; // scanf("%d%d%d%d",&L,&R,&P,&K); Read(L);Read(R);Read(P);Read(K); L^=X;R^=X;P^=X;K^=X; int l=0,r=1e6,ans; while(l<=r)//二分答案 int mid=(l+r)>>1; if(judge(mid)) ans=mid;r=mid-1; else l=mid+1; printf("%d\n",ans); X=ans; return 0;
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