poj3368 Frequent values

思路：

转化为RMQ。

实现：

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 using namespace std;
 5 const int MAXN = 100005;
 6 const int INF = 0x3f3f3f3f;
 7
 8 int a[MAXN], tree[MAXN * 4], n, m, sum[MAXN];
 9
10 void build(int num, int l, int r)
11 {
12     if (l == r) { tree[num] = a[l]; return; }
13     int m = l + r >> 1;
14     build(num * 2, l, m);
15     build(num * 2 + 1, m + 1, r);
16     tree[num] = max(tree[num * 2], tree[num * 2 + 1]);
17 }
18
19 int query(int num, int l, int r, int x, int y)
20 {
21     if (x <= l && y >= r) return tree[num];
22     int ans = -INF, m = l + r >> 1;
23     if (x <= m) ans = max(ans, query(num * 2, l, m, x, y));
24     if (y >= m + 1) ans = max(ans, query(num * 2 + 1, m + 1, r, x, y));
25     return ans;
26 }
27
28 int main()
29 {
30     while (scanf("%d", &n), n)
31     {
32         memset(a, 0, sizeof a);
33         memset(sum, 0, sizeof sum);
34         scanf("%d", &m);
35         int last = -INF, tmp, now = 0;
36         for (int i = 1; i <= n; i++)
37         {
38             scanf("%d", &tmp);
39             if (tmp == last) a[now]++;
40             else last = tmp, a[++now]++;
41         }
42         build(1, 1, now);
43         for (int i = 1; i <= now; i++) sum[i] = sum[i - 1] + a[i];
44         int l, r;
45         for (int i = 0; i < m; i++)
46         {
47             scanf("%d %d", &l, &r);
48             int ls = lower_bound(sum, sum + now + 1, l) - sum;
49             int rs = lower_bound(sum, sum + now + 1, r) - sum;
50             int ans = -1;
51             if (ls == rs) ans = r - l + 1;
52             else
53             {
54                 ans = max(sum[ls] - l + 1, r - sum[rs - 1]);
55                 if (rs - ls > 1) ans = max(ans, query(1, 1, now, ls + 1, rs - 1));
56             }
57             printf("%d\n", ans);
58         }
59     }
60     return 0;
61 }

原文地址：https://www.cnblogs.com/wangyiming/p/8455107.html