poj3368 Frequent values

思路：

转化为RMQ。

实现：

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 using namespace std;
 const int MAXN = ;
 const int INF = 0x3f3f3f3f;

 int a[MAXN], tree[MAXN * ], n, m, sum[MAXN];

 void build(int num, int l, int r)
 {
     if (l == r) { tree[num] = a[l]; return; }
     int m = l + r >> ;
     build(num * , l, m);
     build(num *  + , m + , r);
     tree[num] = max(tree[num * ], tree[num *  + ]);
 }

 int query(int num, int l, int r, int x, int y)
 {
     if (x <= l && y >= r) return tree[num];
     int ans = -INF, m = l + r >> ;
     if (x <= m) ans = max(ans, query(num * , l, m, x, y));
     if (y >= m + ) ans = max(ans, query(num *  + , m + , r, x, y));
     return ans;
 }

 int main()
 {
     while (scanf("%d", &n), n)
     {
         memset(a, , sizeof a);
         memset(sum, , sizeof sum);
         scanf("%d", &m);
         int last = -INF, tmp, now = ;
         for (int i = ; i <= n; i++)
         {
             scanf("%d", &tmp);
             if (tmp == last) a[now]++;
             else last = tmp, a[++now]++;
         }
         build(, , now);
         for (int i = ; i <= now; i++) sum[i] = sum[i - ] + a[i];
         int l, r;
         for (int i = ; i < m; i++)
         {
             scanf("%d %d", &l, &r);
             int ls = lower_bound(sum, sum + now + , l) - sum;
             int rs = lower_bound(sum, sum + now + , r) - sum;
             int ans = -;
             if (ls == rs) ans = r - l + ;
             else
             {
                 ans = max(sum[ls] - l + , r - sum[rs - ]);
                 if (rs - ls > ) ans = max(ans, query(, , now, ls + , rs - ));
             }
             printf("%d\n", ans);
         }
     }
     return ;
 }