数据结构专题——链表处理A1052.Linked List Sorting(25) （静态链表）

自己写的，少了异常处理。

#include <bits/stdc++.h>
#include<math.h>
#include <string>
using namespace std;
const int maxn = 100010;
struct Node{//定义静态链表
    int data;
    int address;
    int next;
    bool flag;
}node[maxn];
bool cmp(Node a,Node b){
    //return a.data < b.data;
    if(a.flag != b.flag){
        return a.flag > b.flag;
    }else{
        return a.data < b.data;
    }
}
int main(){
    int n,addre;
    scanf("%d%d",&n,&addre);
    int add,d,nex;
    for(int i=0;i<n;++i){
        scanf("%d%d%d",&add,&d,&nex);
        node[add].data = d;
        node[add].next = nex;
        node[add].flag = true;
        node[add].address = add;
    }
    sort(node,node+maxn,cmp);
    printf("%d %05d\n",n,node[0].address);
    for(int i =0;i<n;++i){
        if(i != n-1){
            node[i].next = node[i+1].address;
        }else{
            node[i].next = -1;
        }
        if(i != n-1){
            printf("%05d %d %05d\n",node[i].address,node[i].data,node[i].next);
        }else{
            printf("%05d %d %d\n",node[i].address,node[i].data,node[i].next);
        }
    }
    system("pause");
    return 0;
}

由于题目可能会有无效节点，即不在题目给出的首地址开始的链表上

数据里面还有全部是无效的情况，这时就要根据有效节点的个数特判输出“0 -1”

#include <bits/stdc++.h>
#include<math.h>
#include <string>
using namespace std;
const int maxn = 100010;
struct Node{//定义静态链表
    int data;
    int address;
    int next;
    bool flag;
}node[maxn];
bool cmp(Node a,Node b){
    //return a.data < b.data;
    if(a.flag != b.flag){
        return a.flag > b.flag;
    }else{
        return a.data < b.data;
    }
}
int main(){
    for(int i=0;i<maxn;++i){
        node[i].flag = false;
    }
    int n,begin,address;
    scanf("%d%d",&n,&begin);
    for(int i =0;i<n;++i){
        scanf("%d",&address);
        scanf("%d%d",&node[address].data,&node[address].next);
        node[address].address = address;
    }
    int count = 0,p = begin;
    //枚举链表，对flag进行标记,同时计数有效节点个数
    while(p != -1){
        node[p].flag = true;
        count++;
        p = node[p].next;
    }
    if(count == 0){//特判，新链表中没有节点时输出0 -1
        printf("0 -1");
    }else{
        //筛选有效节点，并按data从小到大排序
        sort(node,node+maxn,cmp);
        printf("%d %05d\n",count,node[0].address);
        for(int i=0;i<count;++i){
            if(i != count -1){
                printf("%05d %d %05d\n",node[i].address,node[i].data,node[i+1].address);
            }else{
                printf("%05d %d -1\n",node[i].address,node[i].data);
            }
        }
    }
    system("pause");
    return 0;
}