（重要！） 静态链表
tips
题目可能会有无效结点，即不在题目给出的首地址开始的链表上。因此要先遍历一遍链表，标记出有效结点。 数据里面还有均无效的情况，这时要输出“0 -1”(我想到了这个情况但只输出了0...太无奈了QAQ)
AC代码
#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
#define MAX 100010
struct Node {
int adress;
int key;
int next;
bool in;
}node[MAX];
bool cmp(Node a, Node b) {
if (a.in == false || b.in == false)
return a.in > b.in;
else return a.key < b.key;
}
int main() {
int start, n;
scanf("%d%d", &n, &start);
for (int i = 0;i < MAX;i++)
node[i].in = false;
for (int i = 0;i < n;i++) {
int ad;
scanf("%d", &ad);
node[ad].adress = ad;
scanf("%d%d",&node[ad].key, &node[ad].next);
}
int p = start;
while (p != -1) {
node[p].in = true;
p = node[p].next;
}
sort(node, node + MAX, cmp);
start = node[0].adress;
n = 0;
while (node[n].in == true) {
node[n].next = node[n + 1].adress;
n++;
}
node[n - 1].next = -1;
printf("%d", n);
if (n > 0)printf(" %05d\n", start);
else printf(" -1\n");
for (int i = 0;i < n;i++) {
printf("%05d %d ", node[i].adress, node[i].key);
if (node[i].next == -1)
printf("-1\n");
else printf("%05d\n", node[i].next);
}
return 0;
}
