Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P3554 [POI2013]LUK-Triumphal arch

P3554 [POI2013]LUK-Triumphal arch

首先发现两个性质:

  • $k$ 具有单调性,如果 $k = x$ 可以胜利,那么 $k > x$ 一定可以胜利。因此考虑将问题使用二分转化为判断性问题。
  • $B$ 不会往回走,在树上,向父亲节点走时,新的白点会越来越少。因此 $B$ 会顺着树向下走。

二分后,每次贪心染色。

考虑二分的边界,任何时候,一定要满足答案最小为根节点的儿子的个数,最大为最大的度数。

查看代码
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#include <cstdio>
#include <algorithm>
using namespace std;
const int N = 3e5 + 10, M = 6e5 + 10;
int n, s[N];
int idx, hd[N], nxt[M], edg[M];
void init(int x, int fa)
{
for (int i = hd[x]; ~i; i = nxt[i])
if (edg[i] != fa)
{
init(edg[i], x);
s[x]++;
}
}
int dfs(int x, int fa, int k)
{
int res = s[x] - k;
for (int i = hd[x]; ~i; i = nxt[i])
if (edg[i] != fa)
{
int t = dfs(edg[i], x, k);
if (t > 0)
res += t;
}
return res;
}
void add(int a, int b)
{
nxt[++idx] = hd[a];
hd[a] = idx;
edg[idx] = b;
}
int main()
{
scanf("%d", &n);
for (int i = 1; i <= n; i++)
hd[i] = -1;
for (int i = 1, a, b; i < n; i++)
{
scanf("%d%d", &a, &b);
add(a, b);
add(b, a);
}
init(1, 0);
int l = s[1], r = 0, res;
for (int i = 1; i <= n; i++)
r = max(r, s[i]);
while (l <= r)
{
int mid = l + r >> 1;
dfs(1, 0, mid) <= 0 ? (r = mid - 1, res = mid) : l = mid + 1;
}
printf("%d", res);
return 0;
}