1133 Splitting A Linked List

最新推荐文章于 2024-05-08 21:39:44 发布

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本文介绍了一个使用 C++ 编写的程序，该程序能够读取一组链表节点，并根据节点值将其分为三类：负数、0 到 k 之间的数和其他数。通过将节点推入不同的向量中实现分类，最后按顺序输出这些节点。

#include<iostream>
#include<vector>
using namespace std;
struct 	Node{
	int value;
	int next;
	int address;
}node[100010];
int main()
{
	#ifdef ONLINE_JUDGE
	#else
	freopen("in.txt","r",stdin);
	#endif
	int n,k;
	int f;
	cin>>f>>n>>k;
	vector<Node> a,b,c;
	for(int i=0;i<n;i++){
		int add,nex;
		int data;
		cin>>add>>data>>nex;
		node[add].next=nex;
		node[add].value=data;
		node[add].address=add;
	}
	int p;
	for(p=f;p!=-1;p=node[p].next){
		if(node[p].value<0){
			a.push_back(node[p]);
		}else if(node[p].value<=k&&node[p].value>=0){
			b.push_back(node[p]);
		}else{
			c.push_back(node[p]);
		}
	}
	for(int i=0;i<a.size();i++){
		if(i<=a.size()-2){
			a[i].next=a[i+1].address;
			printf("%05d %d %05d\n",a[i].address,a[i].value,a[i].next);
			//cout<<a[i].address<<' '<<a[i].value<<' '<<a[i].next<<endl;
		}else if(b.size()!=0){
			a[i].next=b[0].address;
			printf("%05d %d %05d\n",a[i].address,a[i].value,a[i].next);
			//cout<<a[i].address<<' '<<a[i].value<<' '<<a[i].next<<endl;
		}else{
			a[i].next=-1;
			printf("%05d %d %d\n",a[i].address,a[i].value,a[i].next);
			//cout<<a[i].address<<' '<<a[i].value<<' '<<a[i].next<<endl;
		}
	}
	for(int i=0;i<b.size();i++){
		if(i<=b.size()-2){
			b[i].next=b[i+1].address;
			printf("%05d %d %05d\n",b[i].address,b[i].value,b[i].next);
			//cout<<b[i].address<<' '<<b[i].value<<' '<<b[i].next<<endl;
		}else if(c.size()!=0){
			b[i].next=c[0].address;
			printf("%05d %d %05d\n",b[i].address,b[i].value,b[i].next);
			//cout<<b[i].address<<' '<<b[i].value<<' '<<b[i].next<<endl;
		}else{
			b[i].next=-1;
			printf("%05d %d %d\n",b[i].address,b[i].value,b[i].next);
			//cout<<b[i].address<<' '<<b[i].value<<' '<<b[i].next<<endl;
		}
	}
	for(int i=0;i<c.size();i++){
		if(i<=c.size()-2){
			c[i].next=c[i+1].address;
			printf("%05d %d %05d\n",c[i].address,c[i].value,c[i].next);
			//cout<<c[i].address<<' '<<c[i].value<<' '<<c[i].next<<endl;
		}else{
			c[i].next=-1;
			printf("%05d %d %d\n",c[i].address,c[i].value,c[i].next);
			//cout<<c[i].address<<' '<<c[i].value<<' '<<c[i].next<<endl;
		}
	}
	return 0;
}