PAT1052 Linked List Sorting

思路：建立链表，并通过排序将该链表的结点移到数组左端。

最后一组数据是建立不出链表的情况

链表排序写法：

bool cmp(node a,node b){
    if(a.flag ==0||b.flag==0) {
        return a.flag>b.flag; //把无效结点排在后边
    }
    return a.num<b.num;
}

#include <iostream>
#include<bits/stdc++.h>
#define rep(i,a,n) for(int i=a;i<n;i++)
#define sca(x) scanf("%d",&x)
#define sca2(x,y) scanf("%d%d",&x,&y)
#define scl(x) scanf("%lld",&x)
#define scs(x) scanf("%s",x)
#define pri(x) printf("%d\n",x)
#define pri2(x,y) printf("%d %d\n",x,y)
#define prs(x) printf("%s\n",(x))
#define prl(x) printf("%lld\n",x)
#define ll long long
#define PII pair<int,int>
#define eps 1e-6
#define inf 1e17
#define INF 0x3f3f3f3f
using namespace std;
const int maxn=1e6+5;
struct node{
    int id,nxt;
    int num;
    int flag;
}a[maxn];
int n;
int id,nxt;
int head,num;
bool cmp(node a,node b){
    if(a.flag ==0||b.flag==0) {
        return a.flag>b.flag;
    }
    return a.num<b.num;
}
int main(){
  cin>>n>>head;
  rep(i,0,n){
    cin>>id>>num>>nxt;
    a[id].id=id;
    a[id].num=num;
    a[id].nxt=nxt;
  }
  int cnt=0;
  for(int i=head;i!=-1;i=a[i].nxt){
    a[i].flag=1;
    cnt++;
  }
  if(cnt==0){
    puts("0 -1");
    return 0;
  }
  sort(a,a+maxn,cmp);
  printf("%d %05d\n",cnt,a[0].id);
  for(int i=0;i<cnt;i++){
    if(i==cnt-1) printf("%05d %d -1\n",a[i].id,a[i].num);
    else printf("%05d %d %05d\n",a[i].id,a[i].num,a[i+1].id);
  }
}