POJ - 2823 Sliding Window: 滑动窗口 单调队列

题目点此跳转

思路

 题目意思是给你一个数组，让你分别求出所有的区间长度为k的区间的最小值和最大值。

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

 从左到右滑动窗口， 每次将窗口右侧加入单调队列(比它大的出队)， 然后左侧不在窗口内的出队，这样保证单调队列首元素即为当前窗口位置的最大值。

代码

#include <algorithm>
#include <iostream>
#include <sstream>
#include <utility>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <cstring>
#include <cstdio>
#include <cmath>
#define met(a,b) memset(a, b, sizeof(a));
#define IN freopen("in.txt", "r", stdin);
#define OT freopen("ot.txt", "w", stdout);
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int maxn = 1e6 + 100;
const LL INF = 0x7fffffff;
const int dir[5][2] = {0,0,-1,0,1,0,0,-1,0,1};
const int MOD = 1e9 + 7;
const double eps = 1e-6;

int n, k, a[maxn], q[maxn], ft, rr;
int pos[maxn];

int main() {
#ifdef _LOCAL
IN; //OT;
#endif // _LOCAL

while(scanf("%d%d", &n, &k) == 2) {
for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);

ft = rr = 0; met(pos, 0);
q[rr++] = a[1]; pos[rr-1] = 1;
for(int i = 2; i <= k; ++i) {
while(rr > ft && q[rr-1] > a[i]) --rr;
q[rr++] = a[i], pos[rr-1] = i;
}
printf("%d ", q[ft]);
for(int i = k+1; i <= n; ++i) {
while(rr > ft && q[rr-1] > a[i]) --rr;
q[rr++] = a[i], pos[rr-1] = i;
if(pos[ft] < i-k+1) ++ft;
printf("%d ", q[ft]);
}
printf("\n");

ft = rr = 0; met(pos, 0);
q[rr++] = a[1]; pos[rr-1] = 1;
for(int i = 2; i <= k; ++i) {
while(rr > ft && q[rr-1] < a[i]) --rr;
q[rr++] = a[i], pos[rr-1] = i;
}
printf("%d ", q[ft]);
for(int i = k+1; i <= n; ++i) {
while(rr > ft && q[rr-1] < a[i]) --rr;
q[rr++] = a[i], pos[rr-1] = i;
if(pos[ft] < i-k+1) ++ft;
printf("%d ", q[ft]);
}
printf("\n");

}

return 0;
}