POJ 3368 Frequent values

标签：poj   acm   

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

以前还没怎么结构体和数组有很大的区别，做完这个题以后，深有体会，结构体在树状数组里面太重要了，这个题打算用线段树和RMQ写。

这个题让我有种树中树的感觉，实现过程在代码中详解！！！

AC代码如下：

///线段树    547MS  3660K

#include<iostream>
#include<cstring>
#include<cstdio>
#define M 100010
using namespace std;

struct H
{
    int l,r,nn;
}trees[M];

struct HH
{
    int l,r,um;
}tree[M];

int num[M],hash[M];
int dp[M][20];
int n,m;

void build_trees(int jd ,int l,int r)
{
    tree[jd].l=l;tree[jd].r=r;
    if(l==r)
    {
        tree[jd].um=trees[l].nn;
        return ;
    }
    int mid = (l+r)/2;
    build_trees(jd*2,l,mid);
    build_trees(jd*2+1,mid+1,r);
    tree[jd].um=max(tree[jd*2].um,tree[jd*2+1].um);
}

int query(int jd,int l,int r)
{
    int ans = 0;
    if(l<=tree[jd].l&&r>=tree[jd].r)
        return tree[jd].um;
    int mid = (tree[jd].l+tree[jd].r)/2;
    if(l<=mid)  ans=max(ans,query(jd*2,l,r));
    if(r>mid)  ans=max(ans,query(jd*2+1,l,r));
    return ans;
}

int main()
{
    int i,j;
    int a,b;
    while(~scanf("%d",&n))
    {
        if(n==0)
            break;
        scanf("%d",&m);
        for(i=1;i<=n;i++)
        {
            scanf("%d",&num[i]);
        }
        memset(trees,0,sizeof trees);
        int tt=1;
        trees[tt].l=1;
        int cont =1;
        hash[1]=1;
        for(i=2;i<=n;i++)
        {
            if(num[i]!=num[i-1])
            {
                trees[tt].r=i-1;
                trees[tt].nn=cont;
                tt++;
                trees[tt].l=i;
                cont=1;
                hash[i]=tt;
            }
            else{
                cont++;hash[i]=tt;
            }
            if(i==n)
            {
                trees[tt].r=i;trees[tt].nn=cont;
            }
        }
//        for(i=1;i<=n;i++)
//            cout<<num[i]<<" ";
//        cout<<endl;
//        for(i=1;i<=n;i++)
//            cout<<hash[i]<<" ";//hash记录的是i号在第几类
//        cout<<endl;
//        for(i=1;i<=tt;i++)
//            cout<<"<"<<trees[i].l<<" "<<trees[i].nn<<" "<<trees[i].r<<">"<<endl;//将这个注释消掉，你应该能看明白很多东西
        build_trees(1,1,tt);//建树，一类别编号为左右区间，相同数的个数为最大值建树
        for(i=1;i<=m;i++)
        {
            scanf("%d%d",&a,&b);
            if(hash[a]==hash[b])//当区间同类的时候，就是区间长度
                printf("%d\n",(b-a+1));
            else//不同类时
            {
                int ans = trees[hash[a]].r-a+1;//计算最左区间
                if(b-trees[hash[b]].l+1>ans)
                    ans=b-trees[hash[b]].l+1;//计算最右区间
                if(hash[b]-hash[a]>1)
                {
                    int aa=query(1,hash[a]+1,hash[b]-1);//如果中间还有不同的类，需要求最值
                    if(aa>ans)
                        ans=aa;
                }
                printf("%d\n",ans);
            }

        }
    }
    return 0;
}

RMQ~~~~！！快得不多，写法也差不多

/// RMQ   454MS  9948K

<span style="font-size:14px;">#include<iostream>
#include<cstring>
#include<cstdio>
#define M 100010
using namespace std;

struct H
{
    int l,r,nn;
}trees[M];

int dp[M][20];

int num[M],hash[M];
int n,m;

void RMQ(int n)
{
    int i,j;
    memset(dp,0,sizeof dp);
    for(i=1;i<=n;i++)
        dp[i][0]=trees[i].nn;
    for(j=1;(1<<j)<=n;j++)
        for(i=1;i+(1<<j)-1<=n;i++)
        dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
}

int rmq(int l,int r)
{
    int k=0;
    while((1<<(k+1))<=r-l+1)
    {
        k++;
    }
    return max(dp[l][k],dp[r-(1<<k)+1][k]);
}

int main()
{
    int i,j;
    int a,b;
    while(~scanf("%d",&n))
    {
        if(n==0)
            break;
        scanf("%d",&m);
        for(i=1;i<=n;i++)
        {
            scanf("%d",&num[i]);
        }
        memset(trees,0,sizeof trees);
        int tt=1;
        trees[tt].l=1;
        int cont =1;
        hash[1]=1;
        for(i=2;i<=n;i++)
        {
            if(num[i]!=num[i-1])
            {
                trees[tt].r=i-1;
                trees[tt].nn=cont;
                tt++;
                trees[tt].l=i;
                cont=1;
                hash[i]=tt;
            }
            else{
                cont++;hash[i]=tt;
            }
            if(i==n)
            {
                trees[tt].r=i;trees[tt].nn=cont;
            }
        }
        RMQ(tt);
//        for(i=1;i<=n;i++)
//            cout<<num[i]<<" ";
//        cout<<endl;
//        for(i=1;i<=n;i++)
//            cout<<hash[i]<<" ";
//        cout<<endl;
//        for(i=1;i<=tt;i++)
//            cout<<"<"<<trees[i].l<<" "<<trees[i].nn<<" "<<trees[i].r<<">"<<endl;
        for(i=1;i<=m;i++)
        {
            scanf("%d%d",&a,&b);
            if(hash[a]==hash[b])
                printf("%d\n",(b-a+1));
            else
            {
                int ans = trees[hash[a]].r-a+1;
                if(b-trees[hash[b]].l+1>ans)
                    ans=b-trees[hash[b]].l+1;
                if(hash[b]-hash[a]>1)
                {
                    int aa=rmq(hash[a]+1,hash[b]-1);
                    if(aa>ans)
                        ans=aa;
                }
                printf("%d\n",ans);
            }

        }
    }
    return 0;
}</span>

POJ 3368 Frequent values,布布扣,bubuko.com

POJ 3368 Frequent values

标签：poj   acm   

原文地址：http://blog.csdn.net/hanhai768/article/details/37907899