POJ 3368 & UVA 11235 - Frequent values-优快云博客

本文链接：https://blog.youkuaiyun.com/Freenm/article/details/71121994

这是一篇关于RMQ(Range Minimum Query)应用的解题报告，详细介绍了如何解决POJ 3368和UVA 11235这两个编程题目，主要探讨了在数据结构和算法中的频数查询问题。

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RMQ应用题。

解题思路参考：http://blog.youkuaiyun.com/libin56842/article/details/46482803

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define MAXN 100000+5
using namespace std;

int num[MAXN],a[MAXN];
int n,q,seg_num;
struct Seg{
	int value,left,right,len;
}seg[MAXN];

int dp[MAXN][20];
void RMQ_init()
{
	for(int i=1;i<=seg_num;i++) dp[i][0]=seg[i].len;
	double j_max=(log(seg_num)/log(2.0));
	for(int j=1;j<=j_max;j++)
	{
		for(int i=1;i<=seg_num;i++)
			if(i+(1 << j)-1 <= seg_num) dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
	}
}
int query(int l,int r)
{
	int k=log(r-l+1)/log(2.0);
	return max(dp[l][k],dp[r-(1<<k)+1][k]);
}

int main()
{
	//freopen("input.txt","r",stdin);
	//freopen("output.txt","w",stdout); 
	while(scanf("%d",&n) && n!=0)
	{
		scanf("%d",&q);
		seg_num=1;
		for(int i=1;i<=n;i++)
		{
			scanf("%d",&a[i]);
			if(i==1)
			{
				seg[seg_num].value=a[1];
				seg[seg_num].len=1;
				seg[seg_num].left=1;
				seg[seg_num].right=1;
				num[i]=seg_num;
			}
			else if(a[i]==a[i-1])
			{
				seg[seg_num].len++;
				seg[seg_num].right++;
				num[i]=seg_num;
			}
			else
			{
				seg_num++;
				seg[seg_num].value=a[i];
				seg[seg_num].left=i;
				seg[seg_num].right=i;
				seg[seg_num].len=1;
				num[i]=seg_num;
			}
		}
		RMQ_init();
		while(q--)
		{
			int l,r;
			scanf("%d%d",&l,&r);
			if(num[l]==num[r])
			{
				printf("%d\n",r-l+1);
			}
			else
			{
				int ans=0;
				if(num[l]+1 <= num[r]-1) ans=query(num[l]+1,num[r]-1);
				ans=max( ans , max( seg[ num[l] ].right - l  + 1 , r - seg[ num[r] ].left + 1 ) );
				printf("%d\n",ans);
			}
		}
	}
}

附一个网上的数据生成器：

#include<iostream>
#include<fstream>
#include<cstdlib>
#include<time.h>
using namespace std;
ofstream fout("input.txt");
int main(){
    
    srand((unsigned)time(NULL));
    
    int n = rand()%100 + 1;
    int q = rand()%100 + 1;
    
    fout<<n<<" "<<q<<endl;
    
    int start = rand()%1000 - 500;
    for(int i =1; i<= n;i++){
        start += rand()%2;
        fout<<start<<" ";
    }
    
    fout<<endl;
    for(int i = 0;i < q;i++){
        start = rand()%n + 1;
        int end = min(rand()%(n - start + 1) + start,n);
        fout<<start<<" "<<end<<endl;
    }
    
    fout<<"0"<<endl;
    return 0;
}