PAT 1074

链接：http://www.patest.cn/contests/pat-a-practise/1074

反转链表，但是PAT上的链表题（其实总共也就两道题）都是可以使诈的，比如list sorting，拿到了数据之后，记录好所有node的val和cur_address，排个序然后再依次打出来就好了... 比起用归并写简单太多也不容易错太多

这道题也是用使诈的方法更不容易错，记录下来每一个节点的值，然后再用栈和vector将结果拿到，注意当n % k == 0时候的情况。

#include <fstream>
#include <iostream>
#include <vector>
#include <string>
#include <stack>

using namespace std;

//#define OJ

#ifdef OJ
#define fin cin
#else
ifstream fin("in.txt");
#endif

struct Node{
	int cur_address;
	int val;
	int next_address;
};

int main(){
	string start;
	int	n, k;
	fin >> start >> n >> k;

	vector<Node> nodes(100000);

	for (int i = 0; i < n; i++){
		Node tmp;
		string s1, s2;

		fin >> s1 >> tmp.val >> s2;
		tmp.cur_address = atoi(s1.c_str());
		tmp.next_address = atoi(s2.c_str());

		nodes[tmp.cur_address] = tmp;
	}

	int start_idx = atoi(start.c_str());
	int cur_address = start_idx;
	vector<int> path;

	while (start_idx != -1){
		path.push_back(start_idx);
		start_idx = nodes[start_idx].next_address;
	}

	vector<int> res;
	int idx = 0;
	n = path.size();
	for (; idx + k - 1 < n; idx += k){
		stack<int> node_stack;
		for (int i = 0; i < k; i++){
			node_stack.push(path[idx + i]);
		}
		while (!node_stack.empty()){
			int tmp = node_stack.top();
			res.push_back(tmp);
			node_stack.pop();
		}
	}
	for (; idx < n; idx++){
		res.push_back(path[idx]);
	}

	for (int i = 0; i < n - 1; i++){
		printf("%05d %d %05d
", nodes[res[i]].cur_address, nodes[res[i]].val, nodes[res[i + 1]].cur_address);
	}
	printf("%05d %d %d
", nodes[res[n - 1]].cur_address, nodes[res[n - 1]].val, -1);

	return 0;
}