HDU Multi-School League 5326 work

HDU Multi-School League 5326 work

Title: http://acm.hdu.edu.cn/showproblem.php?pid=5326

This should be the simplest topic of the two-school game of the week.

Problem-solving idea: It is back to root. Using an array of root[i] = J means that the ancestor of I is J, and this is handled for each input relationship. The result of traversing each number until Root[i] is then 0 (because the root does not have a superior, so the root is 0), and in the process of going back to the root, an array of cnt[i] is used to indicate the number of times I have passed this point. The end is to traverse Cnt[i], add cnt[i] = k results can be AC

#include <bits/stdc++.h>using namespace Std;const int MAX = 100+2;int Root[max] = {};int Cnt[max] = {};//back to root int getr oot (int n, int k) {for    (int i=1; i<=n; ++i)    {        int t = i;        while (Root[t])        {            cnt[root[t]]++;   Each pass the superior, then the superior accumulates, somewhat like the superior reports the same            t = root[t];}    }    int ans = 0;    for (int i=1; i<=n; ++i)    {        if (cnt[i] = = k)            ans++;    }    return ans;} int main (void) {    //freopen ("In.txt", "R", stdin);    int n, K;    while (cin>>n>>k)    {        memset (root, 0, sizeof (root));        memset (CNT, 0, sizeof (CNT));        int u, v;        for (int i=1; i<n; ++i)        {            scanf ("%d%d", &u, &v);            ROOT[V] = u;        }        Back        to the root printf ("%d\n", Getroot (n, k));    }    return 0;}

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HDU Multi-school league 5326 work