UVa 1267 - Network
传送门
题意
一个服务器最多覆盖范围k,求放置最少的服务器覆盖所有叶子。
思路
按距离存入叶子,从距离最大的叶子开始向上数k个祖先放置服务器。把主服务器当成根。
代码
#include <cstdio>
#include <cstring>
#include <vector>
#define LL long long
#define lowbit(x) ((x) & (-x))
const int MAXN = 1000 + 5;
const int INF = 0x3f3f3f3f;
using namespace std;
vector<int> mp[MAXN], nodes[MAXN];
int pa[MAXN], covered[MAXN], k, n;
void Build(int cur, int f, int d)
{
pa[cur] = f;
int no = mp[cur].size();
if (no == 1 && d > k)
nodes[d].push_back(cur);
for (int i = 0; i < no; i++)
{
int v = mp[cur][i];
if (v != f) Build(v, cur, d + 1);
}
}
void DFS(int cur, int f, int d)
{
covered[cur] = 1;
for (int i = 0; i < mp[cur].size(); i++)
{
int v = mp[cur][i];
if (v != f && d < k)
DFS(v, cur, d + 1);
}
}
int Solve()
{
memset(covered, 0, sizeof covered);
int ans = 0;
for (int d = n; d > k; d--)
{
for (int i = 0; i < nodes[d].size(); i++)
{
int t = nodes[d][i];
if (covered[t]) continue;
int v = t;
for (int j = 0; j < k; j++)
v = pa[v];
DFS(v, -1, 0);
ans++;
}
}
return ans;
}
int main()
{
//freopen("input.txt", "r", stdin);
int T, i, j, s;
scanf("%d", &T);
while (T--)
{
scanf("%d%d%d", &n, &s, &k);
for (int i = 1; i <= n; i++)
{
mp[i].clear();
nodes[i].clear();
}
for (i = 0; i < n - 1; i++)
{
int a, b;
scanf("%d%d", &a, &b);
mp[a].push_back(b);
mp[b].push_back(a);
}
Build(s, -1, 0);
printf("%d\n", Solve());
}
return 0;
}