模拟题，注意当k == 1 与 k == n时情况
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#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <map>using
namespace std;const
int N = 100005;struct
Node{ int
pre; int
value; int
lat;}node[N];int
order[N];int
size;map<int, int> pre2idx;void
solve1(int
fir, int
n){ map<int, int>::iterator it = pre2idx.find(fir); order[size++] = it->second; while
(node[it->second].lat != -1) { it = pre2idx.find(node[it->second].lat); order[size++] = it->second; }}void
solve2(int
k, int
n){ int
bound = n / k * k; int
idx = k; while
(idx <= bound) { int
boundt = idx; int
boundb = idx - k + 1; while
(boundt >= boundb) { int
next = 0; if
(boundt > boundb) next = boundt - 1; else { if
(idx < bound) next = idx + k; else
if (idx == bound) { if
(bound < n) next = idx + 1; else
if (bound == n) { printf("%05d %d -1\n", node[order[boundt]].pre, node[order[boundt]].value); break; } } } printf("%05d %d %05d\n", node[order[boundt]].pre, node[order[boundt]].value, node[order[next]].pre); boundt--; } idx += k; } bound++; while
(bound <= n) { if
(node[order[bound]].lat != -1) printf("%05d %d %05d\n", node[order[bound]].pre, node[order[bound]].value, node[order[bound]].lat); else
printf("%05d %d -1\n", node[order[bound]].pre, node[order[bound]].value); bound++; } }int
main(){ int
fir, n, k; while
(scanf("%d%d%d", &fir, &n, &k) != EOF) { pre2idx.clear(); size = 1; for
(int
i = 0; i < n; i++) { scanf("%d%d%d", &node[i].pre, &node[i].value, &node[i].lat); pre2idx.insert(make_pair(node[i].pre, i)); } solve1(fir, n); solve2(k , size - 1); } return
0;}/*00100 6 600000 4 9999900100 1 -168237 6 -133218 3 0000099999 5 6823712309 2 33218*/ |
1074. Reversing Linked List (25),布布扣,bubuko.com
1074. Reversing Linked List (25)
原文：http://www.cnblogs.com/echobfy/p/3576798.html