单调队列典型题
An array of size n ≤ 10 6 is given to you. There is a sliding window of size
k which is moving from the very left of the array to the very right. You can only see the
k numbers in the window. Each time the sliding window moves rightwards by one position. Following is an example:
The array is
[1 3 -1 -3 5 3 6 7], and k is 3.
| Window position | Minimum value | Maximum value |
|---|---|---|
| [1 3 -1] -3 5 3 6 7 | -1 | 3 |
| 1 [3 -1 -3] 5 3 6 7 | -3 | 3 |
| 1 3 [-1 -3 5] 3 6 7 | -3 | 5 |
| 1 3 -1 [-3 5 3] 6 7 | -3 | 5 |
| 1 3 -1 -3 [5 3 6] 7 | 3 | 6 |
| 1 3 -1 -3 5 [3 6 7] | 3 | 7 |
Your task is to determine the maximum and minimum values in the sliding window at each position.
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#define eps 1e-6
#define LL long long
using namespace std;
const int maxn = 2000000;
//const int INF = 0x3f3f3f3f;
int n, k;
int a[maxn];
//单调队列
int qmin[maxn], vmin[maxn], hmin = 1, tmin = 0;
void Min(int a, int i) { //第i个元素a入队
while(hmin<=tmin && vmin[hmin] <= i-k) hmin++; //超范围队首出队
//while(hmin<=tmin && qmin[tmin]>=a) tmin--; //不符合要求队尾出列
int l = hmin, r = tmin;
while(l <= r) {
int m = l+(r-l)/2;
if(qmin[m] >= a) r = m - 1;
else l = m + 1;
}
tmin = ++r;
qmin[tmin] = a;
vmin[tmin] = i;
}
int qmax[maxn], vmax[maxn], hmax = 1, tmax = 0;
void Max(int a, int i) { //第i个元素a入队
while(hmax<=tmax && vmax[hmax] <= i-k) hmax++; //超范围队首出队
//while(hmax<=tmax && qmax[tmax]<=a) tmax--; //不符合要求队尾出列
int l = hmax, r = tmax;
while(l <= r) {
int m = l+(r-l)/2;
if(qmax[m] <= a) r = m - 1;
else l = m + 1;
}
tmax = ++r;
qmax[tmax] = a;
vmax[tmax] = i;
}
int ansMax[maxn], ansMin[maxn];
int main() {
// freopen("input.txt", "r", stdin);
while(scanf("%d%d", &n, &k) == 2) {
hmin = 1, tmin = 0;
hmax = 1, tmax = 0;
for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
for(int i = 1; i < k; i++) {
Min(a[i], i);
Max(a[i], i);
}
for(int i = k; i <= n; i++) {
Min(a[i], i);
ansMin[i-k] = qmin[hmin];
Max(a[i], i);
ansMax[i-k] = qmax[hmax];
}
for(int i = 0; i <= n-k; i++)
if(i != n-k) printf("%d ", ansMin[i]);
else printf("%d\n", ansMin[i]);
for(int i = 0; i <= n-k; i++)
if(i != n-k) printf("%d ", ansMax[i]);
else printf("%d\n", ansMax[i]);
}
return 0;
}
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原文地址：http://blog.csdn.net/u014664226/article/details/47075681
