Frequent values （线段树）

Frequent values （线段树）

You are given a sequence of n integers a1, a2, . . . , an in non-decreasing order. In addition to that, youare given several queries consisting of indices i and j (1 ≤ i ≤ j ≤ n). For each query, determine themost frequent value among the integers ai, . . . , aj .InputThe input consists of several test cases. Each test case starts with a line containing two integers n andq (1 ≤ n, q ≤ 100000). The next line contains n integers a1, . . . , an (−100000 ≤ ai ≤ 100000, for eachi ∈ {1, ..., n}) separated by spaces. You can assume that for each i ∈ {1, . . . , n − 1}: ai ≤ ai+1. Thefollowing q lines contain one query each, consisting of two integers i and j (1 ≤ i ≤ j ≤ n), whichindicate the boundary indices for the query.The last test case is followed by a line containing a single ‘0’.OutputFor each query, print one line with one integer: The number of occurrences of the most frequent valuewithin the given range.Note: A naive algorithm may not run in time!Sample Input10 3-1 -1 1 1 1 1 3 10 10 102 31 105 100Sample Output143

题目大概：

给出n个整数，求区间l，r之间的最多相同元素的数量，这n个整数满足（ai<=aj）当 i<=j 时。

思路：

看到这个题，就想把相同的元素合并成一个点，值为这个元素的数量，求最大相同元素个数，就是求最值。这样就转化为一个简单的线段树求区间最值的问题了。但是这个区间的端点需要特殊考虑一下，因为求得区间不一定是刚好是在每个元素的分界线上。这样，当区间很小时，也需要特殊考虑一下。

代码：

#include#include#include#include#includeusing namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1const int maxn=1e5+10;const int INF=0x3f3f3f3f;int Map[maxn];int a[maxn],b[maxn],c[maxn];int sum[maxn<<3];int cnt;void pushup(int rt){ sum[rt]=max(sum[rt<<1],sum[rt<<1|1]);}void build(int l,int r,int rt){ if(l==r) { sum[rt]=a[++cnt]; return; } int m=(l+r)>>1; build(lson); build(rson); pushup(rt);}int quert(int L,int R,int l,int r,int rt){ if(L<=l&&r<=R) { return sum[rt]; } int m=(l+r)>>1; int ret=0; if(L<=m)ret=max(ret,quert(L,R,lson)); if(R>m)ret=max(ret,quert(L,R,rson)); return ret;}int main(){ int n,m; while(~scanf("%d",&n)&&n) { memset(a,0,sizeof(a)); memset(c,0,sizeof(c)); memset(Map,0,sizeof(Map)); memset(b,0,sizeof(b)); memset(sum,0,sizeof(sum)); scanf("%d",&m); int u; cnt=0; int pre=INF; int ans=0; for(int i=1; i<=n; i++) { scanf("%d",&u); b[i]=u; if(u==pre) { Map[i]=ans; a[ans]++; c[i]=c[i-1]+1; } else { Map[i]=++ans; a[ans]++; c[i]=1; } pre=u; } build(1,ans,1); int l,r; for(int i=1; i<=m; i++) { scanf("%d%d",&l,&r); if(Map[l]==Map[r]) //区间只有一个点 { printf("%d\n",r-l+1); } else if(Map[r]-Map[l]==1)//区间有两个点，不需要用线段树 { printf("%d\n",max((a[Map[l]]-c[l]+1),c[r])); } else //区间有3个点以上，中间的点需要用线段树，两端特殊处理。 { int max_1=quert(Map[l]+1,Map[r]-1,1,ans,1); int max_2=a[Map[l]]-c[l]+1; int max_3=c[r]; int ma=max(max_1,max(max_2,max_3)); printf("%d\n",ma); } } } return 0;}

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