PAT A1097 Deduplication on a Linked List (25分)
测试点2是重复数字的链表为空,此时不需要输出-1,所以需要把seq处理包装起来
而如果第一个非重复链表为空,需要输出-1
#include<cstdio>
#include<vector>
#include<set>
#include<algorithm>
using namespace std;
const int N = 100010;
struct Node{
int address,key,next;
bool isvis = false;
}node[N];
set<int> hasabs;
vector<int> sepadd;//存放不要的地址
int main(){
int n,start;
scanf("%d %d",&start,&n);
int waitaddress;
for(int i = 0;i<n;i++){
int taddress,tkey,tnext;
scanf("%d %d %d",&taddress,&tkey,&tnext);
node[taddress].address = taddress;
node[taddress].key = tkey;
node[taddress].isvis = true;
node[taddress].next = tnext;
}
int count =1;
while(start!=-1){
if(hasabs.find(abs(node[start].key))==hasabs.end()){
if(count!=1) printf("%05d\n",start);//保留5位
hasabs.insert(abs(node[start].key));
printf("%05d %d ",start,node[start].key);
}else{
sepadd.push_back(node[start].address);
}
start = node[start].next;
count++;
}
printf("-1\n");
//处理sep
if(sepadd.size()!=0){
count = 1;
for(int i = 0;i<sepadd.size();i++){
if(count!=1) printf("%05d\n",sepadd[i]);
printf("%05d %d ",sepadd[i],node[sepadd[i]].key);
count++;
}
printf("-1\n");
}
return 0;
}