题目信息
1074. Reversing Linked List (25)
时间限制400 ms
内存限制65536 kB
代码长度限制16000 B
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
解题思路
提取出链表后分组翻转
AC代码
#include <cstdio>
#include <vector>
#include <map>
#include <algorithm>
using namespace std;
struct {
char front[8], next[8];
int value;
}node[100005];
int main()
{
int begin, n, k;
map<int, pair<int, int> > mp;
scanf("%d%d%d", &begin, &n, &k);
for (int i = 0; i < n; ++i){
scanf("%s%d%s", node[i].front, &node[i].value, node[i].next);
mp[atoi(node[i].front)] = make_pair(atoi(node[i].next), i);
}
vector<int> rs;
while (begin != -1){
rs.push_back(mp[begin].second);
begin = mp[begin].first;
}
for (int i = 0; i + k <= rs.size(); i += k){
reverse(rs.begin() + i, rs.begin() + i + k);
}
printf("%s %d ", node[rs[0]].front, node[rs[0]].value);
for (int i = 1; i < rs.size(); ++i){
printf("%s\n%s %d ", node[rs[i]].front, node[rs[i]].front, node[rs[i]].value);
}
printf("-1\n");
return 0;
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