02-線性結構3 Reversing Linked List (25分)

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

1定義一個100000的結構數組a，下標即地址
2數據讀入
3定義一個數組b，按照鏈接順序讀取a中地址加入b
4 反轉b中地址，按照反轉後的順序輸出a中的內容

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;
void rev(vector<int>::iterator, vector<int>::iterator);
struct Node {
	int Data;
	int Next;
};

int main() {
	vector<Node> nodes(100000);
	vector<int> v;
	vector<int>::iterator it;
	int n, head, k, i;
	int addr, data, next;
	scanf("%d%d%d", &head, &n, &k);
	while (n--) {
		scanf("%d%d%d", &addr, &data, &next);
		nodes[addr].Data = data;
		nodes[addr].Next = next;
	}

	addr = head;
	while (addr != -1) {
		v.push_back(addr);
		addr = nodes[addr].Next;
	}

	i = 0;
	n = v.size();  //注意有多餘結點不在鏈表上
	while(k > 1 && i+k <= n){
		//reverse(v.begin() + i, v.begin() + (i + k));
		rev(v.begin() + i, v.begin() + (i + k));
		i = i + k;		
	}

	for (it = v.begin();it != (v.end() - 1);++it)
		printf("%05d %d %05d\n", *it, nodes[*it].Data, *(it + 1));
	printf("%05d %d %d\n", *it, nodes[*it].Data,-1);
	return 0;
}

void rev(vector<int>::iterator b, vector<int>::iterator e) {
	int t;
	e--;
	while (b < e) {
		t = *b;
		*b = *e;
		*e = t;
		++b;--e;
	}
}