1133 Splitting A Linked List (25 分)

Given a singly linked list, you are supposed to rearrange its elements so that all the negative values appear before all of the non-negatives, and all the values in [0, K] appear before all those greater than K. The order of the elements inside each class must not be changed. For example, given the list being 18→7→-4→0→5→-6→10→11→-2 and K being 10, you must output -4→-6→-2→7→0→5→10→18→11.

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 (≤) which is the total number of nodes, and a positive K (≤). The address of a node is a 5-digit nonnegative integer, and NULL is represented by −.

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 in [, and Next is the position of the next node. It is guaranteed that the list is not empty.

Output Specification:

For each case, output in order (from beginning to the end of the list) the resulting linked list. Each node occupies a line, and is printed in the same format as in the input.

Sample Input:

00100 9 10
23333 10 27777
00000 0 99999
00100 18 12309
68237 -6 23333
33218 -4 00000
48652 -2 -1
99999 5 68237
27777 11 48652
12309 7 33218

Sample Output:

33218 -4 68237
68237 -6 48652
48652 -2 12309
12309 7 00000
00000 0 99999
99999 5 23333
23333 10 00100
00100 18 27777
27777 11 -1


map处理一下都好说。

 1 #include <bits/stdc++.h>
 2 #define N 100005
 3 using namespace std;
 4 string st;
 5 int n,m;
 6 struct Node
 7 {
 8     int val;
 9     string s;
10 }an[N],bn[N],cn[N];
11 map<string,Node> mp;
12 int main(){
13     ios::sync_with_stdio(false);
14     cin.tie(0);
15     cout.tie(0);
16     cin >> st >> n >> m;
17     string x,y;
18     int z;
19     for(int i = 0 ; i < n; i++){
20         cin >> x >> z >> y;
21         mp[x]={z,y};
22     }
23     Node node;
24     int l1 = 0, l2 = 0, l3 = 0;
25     while(st != "-1"){
26         node = mp[st];
27         if(node.val < 0){
28             an[l1++] = {node.val, st};
29         }else if(node.val >= 0 && node.val <= m){
30             bn[l2++] = {node.val, st};
31         }else{
32             cn[l3++] = {node.val, st};
33         }
34         st = node.s;
35     }
36     for(int i = 0; i < l2; i++){
37         an[l1+i] = bn[i];
38     }
39     l1 = l1 + l2;
40     for(int i = 0; i < l3; i++){
41         an[l1+i] = cn[i];
42     }
43     l1 += l3;
44     for(int i = 0; i < l1-1; i++){
45         cout << an[i].s<<" "<<an[i].val<<" "<<an[i+1].s<<endl;
46     }
47     cout << an[l1-1].s<<" "<<an[l1-1].val<<" "<<"-1"<<endl;
48     return 0;
49 }