POJ 3368 & UVA 11235 - Frequent values

题目链接：http://poj.org/problem?id=3368

RMQ应用题。

解题思路参考：http://blog.csdn.net/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][];
 void RMQ_init()
 {
     for(int i=;i<=seg_num;i++) dp[i][]=seg[i].len;
     double j_max=(log(seg_num)/log(2.0));
     for(int j=;j<=j_max;j++)
     {
         for(int i=;i<=seg_num;i++)
             if(i+( << j)- <= seg_num) dp[i][j]=max(dp[i][j-],dp[i+(<<(j-))][j-]);
     }
 }
 int query(int l,int r)
 {
     int k=log(r-l+)/log(2.0);
     return max(dp[l][k],dp[r-(<<k)+][k]);
 }

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

     fout<<n<<" "<<q<<endl;

     int start = rand()% - ;
     for(int i =; i<= n;i++){
         start += rand()%;
         fout<<start<<" ";
     }

     fout<<endl;
     for(int i = ;i < q;i++){
         start = rand()%n + ;
         int end = min(rand()%(n - start + ) + start,n);
         fout<<start<<" "<<end<<endl;
     }

     fout<<""<<endl;
     return ;
 }