题目链接
https://www.patest.cn/contests/gplt/L2-026
思路 用一个二维vector 来保存 每个人的子女
然后用BFS 广搜下去，当目前的状态 是搜完的时候
那么此时队列里的人都是最小的一辈 标记一下 CUR 然后 讲答案压入VECTOR 然后排序一下 输出来就可以
AC代码
#include <cstdio> #include <cstring> #include <ctype.h> #include <cstdlib> #include <cmath> #include <climits> #include <ctime> #include <iostream> #include <algorithm> #include <deque> #include <vector> #include <queue> #include <string> #include <map> #include <stack> #include <set> #include <numeric> #include <sstream> #include <iomanip> #include <limits> #define CLR(a) memset(a, 0, sizeof(a)) #define pb push_back using namespace std; typedef long long ll; typedef long double ld; typedef unsigned long long ull; typedef pair <int, int> pii; typedef pair <ll, ll> pll; typedef pair<string, int> psi; typedef pair<string, string> pss; const double PI = 3.14159265358979323846264338327; const double E = exp(1); const double eps = 1e-30; const int INF = 0x3f3f3f3f; const int maxn = 1e5 + 5; const int MOD = 1e9 + 7; vector <int> arr[maxn], ans; int n, Cur; queue <int> q; int Count; void bfs(int cur) { int len = q.size(); for (int i = 0; i < len; i++) { int num = q.front(); q.pop(); vector <int>::iterator it; for (it = arr[num].begin(); it != arr[num].end(); it++) { q.push(*it); Count++; } } if (Count == n) { while (!q.empty()) { int num = q.front(); q.pop(); ans.pb(num); } sort(ans.begin(), ans.end()); Cur = cur + 1; return; } bfs(cur + 1); } int main() { scanf("%d", &n); int vis; int num; for (int i = 1; i <= n; i++) { scanf("%d", &num); if (num != -1) arr[num].pb(i); else vis = i; } if (n == 1) printf("1\n1\n"); else { Count = 1; q.push(vis); bfs(1); printf("%d\n", Cur); vector <int>::iterator it; for (it = ans.begin(); it != ans.end(); it++) { if (it != ans.begin()) printf(" "); printf("%d", *it); } printf("\n"); } }转载于:https://www.cnblogs.com/Dup4/p/9433156.html
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