UVA 11235/HDU 1806/POJ 3368 Frequent values

RMQ.按非降序给出一列数，对于每个询问，回答区间内出现最多的数出现的次数。

先把数据压缩一下，记录每个数字出现的次数，开始以及结束位置。那么询问的区间L，R可以分为三部分：L所在区域，R所在区域，和中间。L所在区域：L到该区间右端点的距离。R所在区域：该区间左端点到R的距离。中间部分就是一个RMQ，求区间最值。当LR在同一区域时，直接算就行。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<string>
#include<queue>
#include<cmath>
///LOOP
#define REP(i, n) for(int i = 0; i < n; i++)
#define FF(i, a, b) for(int i = a; i < b; i++)
#define FFF(i, a, b) for(int i = a; i <= b; i++)
#define FD(i, a, b) for(int i = a - 1; i >= b; i--)
#define FDD(i, a, b) for(int i = a; i >= b; i--)
///INPUT
#define RI(n) scanf("%d", &n)
#define RII(n, m) scanf("%d%d", &n, &m)
#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)
#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)
#define RFI(n) scanf("%lf", &n)
#define RFII(n, m) scanf("%lf%lf", &n, &m)
#define RFIII(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define RFIV(n, m, k, p) scanf("%lf%lf%lf%lf", &n, &m, &k, &p)
#define RS(s) scanf("%s", s)
///OUTPUT
#define PN printf("\n")
#define PI(n) printf("%d\n", n)
#define PIS(n) printf("%d ", n)
#define PS(s) printf("%s\n", s)
#define PSS(s) printf("%s ", s)
#define PC(n) printf("Case %d: ", n)
///OTHER
#define PB(x) push_back(x)
#define CLR(a, b) memset(a, b, sizeof(a))
#define CPY(a, b) memcpy(a, b, sizeof(b))
#define display(A, n, m) {REP(i, n){REP(j, m)PIS(A[i][j]);PN;}}
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1

using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int MOD = 1e9+7;
const int INFI = 1e9 * 2;
const LL LINFI = 1e17;
const double eps = 1e-6;
const int N = 111111;
const int M = N << 2;
const int move[8][2] = {0, 1, 0, -1, 1, 0, -1, 0, 1, 1, 1, -1, -1, 1, -1, -1};

int num
, begin
, end
, d
[30], a
, b
;

void init(int k)
{
FFF(i, 1, k)d[i][0] = num[i];
for(int j = 1; (1 << j) <= k; j++)
for(int i = 1; i + (1 << j) - 1 <= k; i++)
d[i][j] = max(d[i][j - 1], d[i + (1 << (j - 1))][j - 1]);
}

int rmq(int L, int R)
{
int k = 0;
while(1 << (k + 1) <= R - L + 1)k++;
return max(d[L][k], d[R - (1 << k) + 1][k]);
}

int main()
{
//freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);

int n, m, k, x, y, p, q, ans;
while(RI(n), n)
{
RI(m);
k = 1;
CLR(num, 0);
FFF(i, 1, n)
{
RI(a[i]);
if(!k || a[i] != a[i - 1])
{
begin[k] = i;
k++;
}
end[k - 1] = i;
num[k - 1]++;
b[i] = k - 1;
}
init(k - 1);
while(m--)
{
RII(x, y);
if(b[x] == b[y])PI(y - x + 1);
else
{
p = end[b[x]];
q = begin[b[y]];
ans = max(p - x + 1, y - q + 1);;
p = b[x] + 1;
q = b[y] - 1;
if(p <= q)ans = max(ans, rmq(p, q));
PI(ans);
}
}
}
return 0;
}