PAT_(链表处理) 1074 Reversing Linked List 1032 Sharing
1074 Reversing Linked List (25分)
Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K=3, then you must output 3→2→1→6→5→4; if K=4, you must output 4→3→2→1→5→6.
Input Specification:
Each input file contains one test case. For each case, the first line contains the address of the first node, a positive N (≤10^5) which is the total number of nodes, and a positive K (≤N) which is the length of the sublist to be reversed. The address of a node is a 5-digit nonnegative integer, and NULL is represented by -1.
Then N lines follow, each describes a node in the format:
Address Data Next
where Address is the position of the node, Data is an integer, and Next is the position of the next node.
Output Specification:
For each case, output the resulting ordered linked list. Each node occupies a line, and is printed in the same format as in the input.
Sample Input:
00100 6 4
00000 4 99999
00100 1 12309
68237 6 -1
33218 3 00000
99999 5 68237
12309 2 33218
Sample Output:
00000 4 33218
33218 3 12309
12309 2 00100
00100 1 99999
99999 5 68237
68237 6 -1说实话，我觉得这题有点无聊
#include
#include
#include
using namespace std;
const int maxn = 1e5 + 10;
struct Node
{
int add, data, next;
}node[maxn];
void init()
{
for (int i = 0; i < maxn; i++)node[i].add = i;
}
int head, n, k;
vectorlist;
int main()
{
scanf("%d%d%d", &head, &n, &k);
init();
for (int i = 0; i < n; i++)
{
int address;
scanf("%d", &address);
scanf("%d%d", &node[address].data, &node[address].next);
}
int p = head;
while (p != -1)
{
list.push_back(node[p]);
p = node[p].next;
}
int group = list.size() / k;
for (int i = 0; i < group; i++)
{
reverse(list.begin() + i*k, list.begin() + i*k + k);
}
for (int i = 0; i < list.size(); i++)
{
printf("%05d %d ", list[i].add, list[i].data);
if (i != list.size() - 1)printf("%05d", list[i + 1].add);
else printf("-1");
printf("\n");
}
return 0;
}
#include
#include
#include
#include
#include
#include
using namespace std;
const int maxn=100010;
struct node{
int address,data,next;
int order;
node(){
this->order=maxn; //invalid
}
}list[maxn];
bool cmp(node a,node b){
return a.orderi*k;j--){
printf("%05d %d %05d\n",list[j].address,list[j].data,list[j-1].address);
}
//每块最后一个节点的处理
printf("%05d %d ",list[i*k].address,list[i*k].data);
if(i
1032 Sharing (25分)
To store English words, one method is to use linked lists and store a word letter by letter. To save some space, we may let the words share the same sublist if they share the same suffix. For example, loading and being are stored as showed in Figure 1.
思路：用hashset判断链表地址有没有重复，好无聊
#include
#include
#include
#include
#include
#include
#include
using namespace std;
const int maxn=100005;
int num[maxn]; //0 index->next
unordered_set st;
int main(){
int add1,add2,n;
memset(num,-1,sizeof(num));
scanf("%d%d%d",&add1,&add2,&n);
int t1,t2;char c;
for(int i=0;i