7-2 Block Reversing (25分)
Given a singly linked list L. Let us consider every K nodes as a block (if there are less than K nodes at the end of the list, the rest of the nodes are still considered as a block). Your job is to reverse all the blocks in L. For example, given L as 1→2→3→4→5→6→7→8 and K as 3, your output must be 7→8→4→5→6→1→2→3.
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 size of a block. 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 8 3
71120 7 88666
00000 4 99999
00100 1 12309
68237 6 71120
33218 3 00000
99999 5 68237
88666 8 -1
12309 2 33218
Sample Output:
71120 7 88666
88666 8 00000
00000 4 99999
99999 5 68237
68237 6 00100
00100 1 12309
12309 2 33218
33218 3 -1
思路：
先遍歷一遍並標記，然後從大到小排序，之後再按照題目分塊從小到大排序。
注意：最後的“尾巴”處理，可能不是整數
Code:
#include<cstdio>
#include<iostream>
#include<vector>
#include<map>
#include<string>
#include<set>
#include<cctype>
#include<algorithm>
using namespace std;
const int INF = 0x3fffffff;
const int maxn = 100010;
struct node{
int address,data,next,flag;
node(){
flag = -1;
}
}Node[maxn];
bool cmp1(node a,node b){
return a.flag > b.flag;
}
bool cmp2(node a,node b){
return a.flag < b.flag;
}
int main(){
int start,n,k;
scanf("%d%d%d",&start,&n,&k);
for(int i = 0; i < n; i++){
int add;
scanf("%d",&add);
scanf("%d%d",&Node[add].data,&Node[add].next);
Node[add].address = add;
}
int cnt = 0;
for(int p = start; p != -1; p = Node[p].next){
Node[p].flag = cnt++;
}
sort(Node,Node+maxn,cmp1);
int temp = cnt % k;
if(temp != 0){
sort(Node,Node+temp,cmp2);
}
for(int i = temp; i + k <= cnt; i+=k){
sort(Node+i,Node+i+k,cmp2);
}
for(int i = 0; i < cnt; i++){
if(i != cnt - 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);
}
}