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

这题是链表题，我们需要用一个vector模拟链表，将小于链表的数字放在前面，大于链表的数字放在后面，我们定义3个vector即可。最后按序输出即可

#include <iostream>
#include <unordered_map>
#include <vector>
#include <deque>
using namespace std;
struct node {
    int addr, val, next;
}tmp;
unordered_map<int, node> m;
vector<node> v, bef, aft;
deque<node> deq;
int main() {
    int start, N, K;
    cin >> start >> N >> K;
    while(N--) {
        cin >> tmp.addr >> tmp.val >> tmp.next;
        m[tmp.addr] = tmp;
    }
    while(start != -1) {
        v.push_back(m[start]);
        start = m[start].next;
    }
    for(auto& x: v) {
        if(x.val > K) aft.push_back(x);
        else if(x.val < 0) bef.push_back(x);
        else deq.push_back(x);
    }
    for(auto& x: aft) deq.push_back(x);
    for(auto& x: deq) bef.push_back(x);
    printf("%05d %d", bef[0].addr, bef[0].val);
    for(int i = 1; i < bef.size(); i++)
        printf(" %05d\n%05d %d", bef[i].addr, bef[i].addr, bef[i].val);
    printf(" -1\n");
    return 0;
}