POJ

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;}