HDU 5326 Company Personnel Management tree problem (multi-school)-Simple DP

Test instructions: Give a company's personnel management tree, for direct and indirect subordinate total K people have how many.

Analysis: Just start a look at the tree is frightened, do not dare to do, then bite the tooth a think, not difficult ah, G elder brother Said is DP, I used I do not know the method, too. My method is to use Vis[i][j] to record I and J Direct affiliation, and then through Vis[i][j] and vis[j][k] and can find vis[i][k] indirect affiliation, A[I]+=A[J],A[I]+=A[K] can. Believe in yourself.

This is the G-Brother code:

Recursive Dafa must learn to use it.

#include <cstdio> #include <cstring> #include <cmath> #include <cstdlib> #include <iostream > #include <algorithm> #include <vector> #include <map> #include <queue> #include <stack& Gt  #include <string> #include <map> #include <set> #define EPS 1e-6 #define LL Long long using namespace std; const int MAXN = + 5;//const int INF = 0x3f3f3f3f;int d[105];vector<int> g[105];int N, k;int son[105];int DP (    int i) {if (d[i]! =-1) return d[i];    D[i] = 0;    for (int j = 0; J < G[i].size (); j + +) {D[i] + = DP (G[i][j]) +1; } return d[i];}    int main () {//Freopen ("Input.txt", "R", stdin);        while (scanf ("%d%d", &n, &k) = = 2) {memset (son, 0, sizeof (son));        memset (d,-1, sizeof (d));        for (int i = 1; I <= n; i++) g[i].clear ();            for (int i = 0; i < n-1; i++) {int u, v;            cin >> u >> v;        G[u].push_back (v); } inT-ans = 0;        for (int i = 1; I <= n; i++) {if (DP (i) = = k) ans++;    } cout << ans << endl; } return 0;}

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HDU 5326 Company Personnel Management tree problem (multi-school)-Simple DP