Frequent values
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 15229 | Accepted: 5550 |
Description
You are given a sequence of n integers a1 , a2 , ... , an in non-decreasing order. In addition to that, you are given several queries consisting of indices i and j (1 ≤ i ≤ j ≤ n). For each query, determine the most frequent value among the integers ai , ... , aj.
Input
The input consists of several test cases. Each test case starts with a line containing two integers n and q (1 ≤ n, q ≤ 100000). The next line contains n integers a1 , ... , an (-100000 ≤ ai ≤ 100000, for each i ∈ {1, ..., n}) separated by spaces. You can assume that for each i ∈ {1, ..., n-1}: ai ≤ ai+1. The following q lines contain one query each, consisting of two integers i and j (1 ≤ i ≤ j ≤ n), which indicate the boundary indices for the
query.
The last test case is followed by a line containing a single 0.
Output
For each query, print one line with one integer: The number of occurrences of the most frequent value within the given range.
Sample Input
10 3 -1 -1 1 1 1 1 3 10 10 10 2 3 1 10 5 10 0
Sample Output
1 4 3
Source
题意：
给你一个非递减序列。有m次询问，每次询问区间之间出现最多的数字出现的次数。
题解：
RMQ/线段树
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#include<cmath>
#include<map>
using namespace std ;
typedef long long ll;
#define inf 100000
inline ll read()
{
ll 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;
}
//******************************************************************
struct ss{
int l,r,num,v;
}a[100001];
int counts[100001];
int dp[100001][63],n,m,p;
void rmq_init()
{
memset(dp,0,sizeof(dp));
for(int i=1;i<=p;i++){
dp[i][0]=counts[i];
}
int k=floor(log((double)n+1)/log(2.0));
for(int j=1;j<=k;j++)
{
for(int i=1;i+(1<<j)-1<=p;i++)
{
int oo=i+(1<<(j-1));
dp[i][j]=max(dp[i][j-1],dp[oo][j-1]);
}
}
}
int RMQ(int x,int y)
{
int oo=floor(log((double)y-x+1)/log(2.0));
return max(dp[x][oo],dp[y-(1<<(oo))+1][oo]);
}
void work()
{
int l,r;
for(int i=1;i<=m;i++)
{
scanf("%d%d",&l,&r);
if(a[l].num==a[r].num){
cout<<r-l+1<<endl;
}
else {
int ans=a[l].r-l+1;
ans=max(r-a[r].l+1,ans);
l=a[l].num+1;
r=a[r].num-1;
if(l<=r) ans=max(ans,RMQ(l,r));
cout<<ans<<endl;
}
}
}
void init()
{
int f=0,r=0,l=0;
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i].v);
if(f&&a[i].v!=a[i-1].v)
{
counts[p]=r-l+1;
for(int j=l;j<=r;j++)
a[j].num=p,a[j].l=l,a[j].r=r;
r++;
p++;
l=r;
}
else if(f){
r++;
}
if(f==0)
{
l=1;
r=1,f=1;
p=1;
}
}
counts[p]=r-l+1;
for(int j=l;j<=r;j++)
a[j].num=p,a[j].l=l,a[j].r=r;
// for(int i=1;i<=n;i++)cout<<counts[i]<<" ";
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
if(n==0)break;
init();
rmq_init();
work();
}
return 0;
}
RMQ
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#include<cmath>
#include<map>
using namespace std ;
typedef long long ll;
#define inf 100000
inline ll read()
{
ll 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;
}
//******************************************************************
struct ss
{
int l,r,v,num;
}tr[100001*7],a[500001];
int counts[200001];
int n,m,p;
void build(int k,int s,int t)
{
tr[k].l=s;
tr[k].r=t;
if(s==t){
tr[k].v=counts[s];
return ;
}
int mid=(s+t)>>1;
build(k<<1,s,mid);
build(k<<1|1,mid+1,t);
tr[k].v=max(tr[k<<1].v,tr[k<<1|1].v);
}
int ask(int k,int s,int t)
{
if(s==tr[k].l&&t==tr[k].r)
{
return tr[k].v;
}
int mid=(tr[k].l+tr[k].r)>>1;
if(t<=mid)return ask(k<<1,s,t);
else if(s>mid)return ask(k<<1|1,s,t);
else {
return max(ask(k<<1,s,mid),ask(k<<1|1,mid+1,t));
}
}
void init()
{
int f=0,l=0,r=0;
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i].v);
if(f&&a[i].v!=a[i-1].v)
{
counts[p]=r-l+1;
for(int j=l;j<=r;j++)a[j].l=l,a[j].r=r,a[j].num=p;
r++;
l=r;
p++;
}
else if(f){
r++;
}
if(!f)
{
f=l=1;
r=p=1;
}
}
counts[p]=r-l+1;
for(int j=l;j<=r;j++)a[j].l=l,a[j].r=r,a[j].num=p;
///for(int i=1;i<=p;i++)cout<<counts[i]<<" ";
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
p=0;
if(n==0)break;
init();
build(1,1,p);
// cout<<p<<endl;
int x,y;
for(int i=1;i<=m;i++)
{
scanf("%d%d",&x,&y);
if(a[x].num==a[y].num){
cout<<y-x+1<<endl;
}
else {
int ans=a[x].r-x+1;
ans=max(ans,y-a[y].l+1);
x=a[x].num+1;
y=a[y].num-1;
if(x<=y)ans=max(ask(1,x,y),ans);
cout<<ans<<endl;
}
}
}
return 0;
}
线段树