POJ 3264 Balanced Lineup (RMQ,ST)

题目链接:
POJ 3264 Balanced Lineup (RMQ,ST)

题意分析:
询问区间$[L,R]$最大值和最小值的差值为多少?

解题思路:
RMQ的ST算法登场。$dpl[i][j]$代表:从第i个数开始,包含第i个数,区间长度为$2^j$范围内的最小值。有转移:$$dpl[i][j] = min(dp[i][j - 1], dp[i + 2^{j - 1}][j - 1])$$
$dph[i][j]$同理。
最终求区间中的最小值,那么我们只需查询:
$k = log_2(r - l + 1)$
$ans = min(dpl[i][k], dpl[i + (1 << k) + 1][k])$

个人感受:
头一次写ST算法,原来也不难啊……以前以为很复杂Orz

具体代码如下:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
#include<algorithm>
#include<cctype>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iomanip>
#include<iostream>
#include<map>
#include<queue>
#include<set>
#include<sstream>
#include<stack>
#include<string>
#define lowbit(x) (x & (-x))
#define ll long long
#define pr(x) cout << #x << " = " << (x) << '\n';
using namespace std;
const int INF = 0x7f7f7f7f;
const int MAXN = 5e4 + 111;
int dpl[MAXN][20], dph[MAXN][20], a[MAXN];
int getMax(int l, int r) {
int k = log2(r - l + 1);
return max(dph[l][k], dph[r - (1 << k) + 1][k]);
}
int getMin(int l, int r) {
int k = log2(r - l + 1);
return min(dpl[l][k], dpl[r - (1 << k) + 1][k]);
}
int main()
{
int n, q;
while (~scanf("%d%d", &n, &q)) {
for (int i = 1; i <= n; ++i) {
scanf("%d", &a[i]);
dpl[i][0] = dph[i][0] = a[i];
}
for (int j = 1; j < 20; ++j) {
for (int i = 1; i <= n; ++i) {
if (i + (1 << j) - 1 <= n) {
dpl[i][j] = min(dpl[i][j - 1], dpl[i + (1 << (j - 1))][j - 1]);
dph[i][j] = max(dph[i][j - 1], dph[i + (1 << (j - 1))][j - 1]);
}
else break;
}
}
int l, r;
for (int i = 0; i < q; ++i) {
scanf("%d%d", &l, &r);
printf("%d\n", getMax(l, r) - getMin(l, r));
}
}
return 0;
}


广告时间

Java学习网站: how2j

VPS: VPS