因为序列非降,所以相等的元素聚在一起,可以使用游程编码。
每一段连续相等的元素都被映射到一个码元 [a0a1…an-1] -> c1
code是码元序列
len是码的长度
left是码的左边界
right是码的右边界
map是序列中的位置到码的映射
0 1 2 3 4 5 6
序列:-1,1,1,2,2,2,4
code: 0 1 2 3
left: 0,1,3,6
right:0,2,5,6
map: 0,1,1,2,2,2,3
任意一段序列可能有三种情况:(小写字母表示码的编号)
1) [x..x]
2) [x..x][y..y]
3) [x..x][完整连续的多段][z..z] (2是3的特殊情况)
当查询一个区间(L,R)
计算出两端的长度,对于中间的区间,就是一个求最大值问题了
当为情况1时,直接等于 R – L + 1
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <cassert>
#include <algorithm>
#include <cmath>
#include <limits>
#include <set>
#include <time.h>
using namespace std;
#define MIN(a, b) a < b ? a : b
#define MAX(a, b) a > b ? a : b
#define F(i, n) for (int i=0;i<(n);++i)
#define REP(i, s, t) for(int i=s;i<=t;++i)
#define UREP(i, s, t) for(int i=s;i>=t;--i)
#define REPOK(i, s, t, o) for(int i=s;i<=t && o;++i)
#define MEM0(addr, size) memset(addr, 0, size)
#define LBIT(x) x&-x
#define PI 3.1415926535897932384626433832795
#define HALF_PI 1.5707963267948966192313216916398
#define eps 1e-15
#define MAXN 100000
#define MAXM 1000
#define MOD 20071027
typedef long long LL;
const double maxdouble = numeric_limits<double>::max();
const int INF = 0x7FFFFFFF;
const int maxlog = 20;
struct RMQ {
int d[MAXN + 1][maxlog];
void init(const vector<int>& A)
{
int n = A.size();
for (int i=0;i<n;++i)
d[i][0] = A[i];
for (int j=1; (1<<j) <= n;++j)
for (int i=0;i + (1<<j) - 1 < n;++i)
d[i][j] = max(d[i][j-1], d[i+(1<<(j-1))][j-1]);
}
int query(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]);
}
};
vector<int> vec;
int num[MAXN + 1];
int _left[MAXN + 1];
int righ[MAXN + 1];
RMQ rmq;
int main()
{
freopen("input.in", "r", stdin);
int n, q;
int last, now, cnt, s, t, index;
while(scanf("%d%d",&n, &q) == 2) {
cin >> last;
cnt = 1;
s = 0;
index = 0;
num[0] = 0;
vec.clear();
REP(i, 1, n-1) {
cin >> now;
if (now != last) {
t = i-1;
vec.push_back(cnt);
_left[index] = s;
righ[index] = t;
last = now;
cnt = 1;
s = i;
++index;
} else
++cnt;
num[i] = index;
}
t = n-1;
vec.push_back(cnt);
_left[index] = s;
righ[index] = t;
rmq.init(vec);
F(i, q) {
int l, r, ans, id_l, id_r;
cin >> l >> r;
--l;
--r;
id_l = num[l];
id_r = num[r];
if (id_l == id_r)
ans = r-l+1;
else {
ans = max(righ[id_l]-l+1, r-_left[id_r]+1);
if (id_r-id_l > 1)
ans = max(ans, rmq.query(id_l+1, id_r-1));
}
printf("%d\n", ans);
}
}
return 0;
}