HDU 6215 Brute Force Sorting（模拟链表思维）

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 1043 Accepted Submission(s): 272

Problem Description

Beerus needs to sort an array of

. If the new array were still unsorted, Beerus would do it again.
Help Beerus predict the final array.

Input

The first line of input contains an integer

Output

For eact test case output two lines.
The first line contains an integer

and the second line should be an empty line.

Sample Input

5 5 1 2 3 4 5 5 5 4 3 2 1 5 1 2 3 2 1 5 1 3 5 4 2 5 2 4 1 3 5

Sample Output

5 1 2 3 4 5 0 2 1 2 2 1 3 3 2 3 5

Source

2017 ACM/ICPC Asia Regional Qingdao Online

【题意】给你一个序列，每次扫一遍序列，同时删除不符合排序标准的数。如果一个数符合标准，则它大于等于左边的数，小于等于右边的数，删完之后右边的数依次填补空位，求删完之后剩下的序列。

【分析】假设我们已经知道了上一次删除的数的位置，那么这一次我们要删的数肯定是上一次删除的数的前驱后缀中选，那么依次模拟链表删下去就好了，复杂度O(N).

#include <bits/stdc++.h>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
#define pb push_back
#define mp make_pair
#define rep(i,l,r) for(int i=(l);i<=(r);++i)
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int N = 1e5+5;;
const int M = 17;
const int mod = 1e9+7;
const int mo=123;
const double pi= acos(-1.0);
typedef pair<int,int>pii;
int n;
int a[N],pre[N],suf[N],pos[N],poss[N];
bool del[N];
int main() {
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        pre[0]=0;suf[n]=n;
        int cnt=0;
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            pre[i]=i-1;
            suf[i]=i+1;
            del[i]=false;
        }
        a[0]=-1;a[n+1]=N;
        for(int i=1;i<=n;i++){
            if(a[i]>=a[i-1]&&a[i]<=a[i+1])continue;
            pos[++cnt]=i;
            del[i]=true;
            int x=suf[i],y=pre[i];
            pre[x]=y;
            suf[y]=x;
        }
        while(cnt){
            int _cnt=0;
            for(int i=1;i<=cnt;i++){
                int l=pre[pos[i]];
                int r=suf[l];
                if(a[r]<a[l]){
                    if(!del[l]){
                        poss[++_cnt]=l;
                        del[l]=true;
                        int x=suf[l],y=pre[l];
                        pre[x]=y;suf[y]=suf[x];
                    }
                    if(!del[r]){
                        poss[++_cnt]=r;
                        del[r]=true;
                        int x=suf[r],y=pre[r];
                        pre[x]=y;suf[y]=x;
                    }
                }
            }
            cnt=_cnt;
            for(int i=cnt;i>=1;i--){
                pos[i]=poss[i];
            }
        }
        int ans=0;
        for(int i=1;i<=n;i++)if(!del[i])ans++;
        printf("%d\n",ans);
        for(int i=1;i<=n;i++)if(!del[i])printf("%d ",a[i]);
        printf("\n");
    }
    return 0;
}