POJ 3368 Frequent values

Frequent values

Time Limit: 2000MS

Memory Limit: 65536K

Total Submissions: 13051

Accepted: 4792

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

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 100010using 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 100010using 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>