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| Time Limit: 12000MS | Memory Limit: 65536K | |
| Total Submissions: 43129 | Accepted: 12740 | |
| Case Time Limit: 5000MS | ||
| 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.
8 3
1 3 -1 -3 5 3 6 7
-1 -3 -3 -3 3 3
3 3 5 5 6 7
题解：打了个线段树，维护区间最大最小值，时间复杂度是O((n - k + 1)·logn);大约六秒半跑过。。友情提示：求区间最大最小时一起求。。。。不然复杂度*常数2，T成狗。。估计也就我这样的才分开求。。
另外，网上的题解说的是O(n)的单调队列，表示现在还并不会。。（这人太弱了吧。。这都不会。。）
CODE:
#include <cstdio> #define REP(i, s, n) for(int i = s; i <= n; i ++) #define REP_(i, s, n) for(int i = n; i >= s; i --) #define max(a, b) a > b ? a : b #define min(a, b) a < b ? a : b #define MAX_N 1000000 + 10 int n, k, data[MAX_N]; struct node{ int l, r, mx, mn; }a[MAX_N << 4]; void updata(int i){ int t1 = i << 1, t2 = t1 + 1; a[i].mx = max(a[t1].mx, a[t2].mx); a[i].mn = min(a[t1].mn, a[t2].mn); } void maketree(int i, int l, int r){ a[i].l = l; a[i].r = r; if(a[i].l == a[i].r){ a[i].mx = a[i].mn = data[l]; return; } int mid = (a[i].l + a[i].r) >> 1, t1 = i << 1, t2 = t1 + 1; maketree(t1, l, mid); maketree(t2, mid + 1, r); updata(i); } int mi, ma; void Query(int i, int l, int r){ if(a[i].l >= l && a[i].r <= r){ mi=min(mi,a[i].mn); ma=max(ma,a[i].mx); return; } if(a[i].l == a[i].r) return; int mid = (a[i].l + a[i].r) >> 1, t1 = i << 1, t2 = t1 + 1; if(r <= mid) Query(t1, l, r); else if(l > mid) Query(t2, l, r); else Query(t1, l, mid), Query(t2, mid + 1, r); } int res_max[MAX_N], res_min[MAX_N]; int main(){ scanf("%d%d", &n, &k); REP(i, 1, n) scanf("%d", &data[i]); maketree(1, 1, n); if(k >= n){ mi = 99999999, ma = -99999999; Query(1, 1, n); printf("%d\n%d\n", mi, ma); } else{ REP(i, 1, n - k + 1){ mi = 99999999, ma = -99999999; Query(1, i, i + k - 1); res_max[i] = ma, res_min[i] = mi; } REP(i, 1, n - k + 1) printf("%d ", res_min[i]); printf("\n"); REP(i, 1, n - k + 1) printf("%d ", res_max[i]); printf("\n"); } return 0; }
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原文地址：http://www.cnblogs.com/ALXPCUN/p/4551480.html
