02-线性结构3 Reversing Linked List (25分)

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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定义一个100000的结构数组a，下标即地址
2数据读入
3定义一个数组b，按照链接顺序读取a中地址加入b
4 反转b中地址，按照反转后的顺序输出a中的内容

#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();  //注意有多余结点不在链表上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;}}

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