73 lines
1.8 KiB
Markdown
73 lines
1.8 KiB
Markdown
## [$POJ$ $1392$-$Ouroboros$ $Snake$](http://poj.org/problem?id=1392)
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**前置题目**
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**[$hihoCoder$ $1182$ 欧拉路·三](https://www.cnblogs.com/littlehb/p/17611667.html)**
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**【兹鼓欧拉回路】**
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这道题和上面那道题几乎一样, 算是变形题吧, 这道题要求构造的$01$字符串就是必须是字典序最小的,
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在上面那道题的注意下建边的顺序即可. 因为是链式前向星法, 应该大边在前。
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#### $Code$
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```cpp {.line-numbers}
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#include <iostream>
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#include <algorithm>
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#include <queue>
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#include <map>
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#include <cstring>
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#include <vector>
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#include <stack>
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#include <cstdio>
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using namespace std;
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const int N = 1 << 16, M = N;
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int len;
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int rl, res[N];
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int st[N];
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// 链式前向星
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int e[M], h[N], idx, w[M], ne[M];
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void add(int a, int b, int c = 0) {
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e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
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}
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void dfs(int u) {
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for (int i = h[u]; ~i; i = h[u]) {
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int id = w[i];
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if (st[id]) continue;
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st[id] = 1;
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h[u] = ne[i]; // 删边的套路优化
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dfs(e[i]);
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res[rl++] = id;
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}
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}
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int main() {
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#ifndef ONLINE_JUDGE
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freopen("POJ1392.in", "r", stdin);
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#endif
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int n, k;
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while (scanf("%d%d", &n, &k), n + k) {
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idx = rl = 0;
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memset(h, -1, sizeof h);
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memset(st, 0, sizeof st);
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memset(res, 0, sizeof res);
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len = 1 << (n - 1);
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for (int a = 0; a < len; a++) {
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int c = a << 1 | 1;
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int b = c % len;
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add(a, b, c);
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c = a << 1;
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b = c % len;
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add(a, b, c);
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}
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dfs(0);
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// 输出这个序列生成的第K个数字是多少？
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printf("%d\n", res[rl - 1 - k]);
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}
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return 0;
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}
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``` |