poj 3368 rmq ***

题意：给出n个数和Q个询问（l,r)，对于每个询问求出(l,r)之间连续出现次数最多的次数。
 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<cstring>
 5 #include<cmath>
 6 #include<queue>
 7 #include<map>
 8 using namespace std;
 9 #define MOD 1000000007
10 const int INF=0x3f3f3f3f;
11 const double eps=1e-5;
12 typedef long long ll;
13 #define cl(a) memset(a,0,sizeof(a))
14 #define ts printf("*****
");
15 int n,m,tt;
16 const int MAXN = 100005;
17 int dp[MAXN][20];
18 int mm[MAXN];
19 //初始化RMQ, b数组下标从1开始，从0开始简单修改
20 void initRMQ(int n,int b[])
21 {
22 mm[0] = -1;
23 for(int i = 1; i <= n;i++)
24 {
25 mm[i] = ((i&(i-1)) == 0)?mm[i-1]+1:mm[i-1];
26 dp[i][0] = b[i];
27 }
28 for(int j = 1; j <= mm[n];j++)
29 for(int i = 1;i + (1<<j) -1 <= n;i++)
30 dp[i][j] = max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
31 }
32 //查询最大值
33 int rmq(int x,int y)
34 {
35 int k = mm[y-x+1];
36 return max(dp[x][k],dp[y-(1<<k)+1][k]);
37 }
38 int a[MAXN],f[MAXN];
39 int main()
40 {
41     int i,j,k;
42     #ifndef ONLINE_JUDGE
43     freopen("1.in","r",stdin);
44     #endif
45     while(scanf("%d",&n)!=EOF)
46     {
47         if(n==0)  break;
48         scanf("%d",&m);
49         for(i=1;i<=n;i++)
50         {
51             scanf("%d",&a[i]);
52             f[1]=1;
53             if(i!=1)
54             {
55                 if(a[i]==a[i-1])
56                 {
57                     f[i]=f[i-1]+1;
58                 }
59                 else
60                     f[i]=1;
61             }
62         }
63         /*for(i=1;i<=n;i++)
64         {
65             printf("%d ",f[i]);
66         }
67         printf("
");*/
68         initRMQ(n,f);
69         while(m--)
70         {
71             int uu,v;
72             scanf("%d%d",&uu,&v);
73             int u=uu;
74             while(u<=v&&a[u]==a[u-1])
75             {
76                 u++;
77             }
78             int ans=rmq(u,v);
79             ans=max(u-uu,ans);
80             printf("%d
",ans);
81         }
82     }
83 }