标签：des style blog http io ar os sp for
| Time Limit: 12000MS | Memory Limit: 65536K | |
| Total Submissions: 40443 | Accepted: 11950 | |
| Case Time Limit: 5000MS | ||
Description
| 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.
Input
Output
Sample Input
8 3 1 3 -1 -3 5 3 6 7
Sample Output
-1 -3 -3 -3 3 3 3 3 5 5 6 7
Source
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<queue>
#include<cstdlib>
#include<vector>
#include<set>
#include<queue>
using namespace std;
int a[1000000];
int aa[1000000];
int bb[1000000];
int n,s;
struct cmp1
{
bool operator()(const int a1,const int a2)
{
return a[a1]>a[a2];
}
};
struct cmp2
{
bool operator()(const int a1,const int a2)
{
return a[a1]<a[a2];
}
};
priority_queue<int,vector<int>,cmp1>q1;
priority_queue<int,vector<int>,cmp2>q2;
int main()
{
scanf("%d%d",&n,&s);
int i,cnt1=0,cnt2=0;
for(i=0;i<n;i++)
scanf("%d",&a[i]);
for(i=0;i<s;i++)
{
q1.push(i);
q2.push(i);
}
aa[cnt1++]=a[q1.top()];
bb[cnt2++]=a[q2.top()];
for(i=s;i<n;i++)
{
q1.push(i);
q2.push(i);
while(i-q1.top()>=s)
q1.pop();
aa[cnt1++]=a[q1.top()];
while(i-q2.top()>=s)
q2.pop();
bb[cnt2++]=a[q2.top()];
}
printf("%d",aa[0]);
for(i=1;i<n-s+1;i++)
printf(" %d",aa[i]);
printf("\n");
printf("%d",bb[0]);
for(i=1;i<n-s+1;i++)
printf(" %d",bb[i]);
printf("\n");
return 0;
}
标签：des style blog http io ar os sp for
原文地址：http://www.cnblogs.com/a972290869/p/4099525.html
