說是樹形DP,其實就是在樹上亂搞。。。。
k=1的時候,随意啦。。。想咋做就咋做,我是用了兩次bfs求出來最長鍊
k=2的時候,用了貪心的思想,顯然如果已經加入一條最長鍊以後,第二條最長鍊有可能不是最優解。。。大概就那個意思
是以我們可以在第一次求完最長鍊以後,把鍊上的所有邊權 都從1改成-1
然後dfs求一次最長路
當然最後都累加到ans裡
答案就是2*(n-1)-ans+k
ps:今天上午剛剛學的c語言程式設計風格啊有沒有感覺很帥。。。。!!!
就是空格太多了。。。。
不知道在比賽前一個星期改程式設計習慣會使什麼結果?
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<queue>
#define rep (i, j, k) for (int i = j; i <= k; i++)
#define MAX 100010
using namespace std;
int to[MAX * 2], next[MAX * 2], head[MAX], value[MAX * 2], n, k, done[MAX], d[MAX];
int num = 1, ans = 0;
int f[MAX], tool, Max = 0;
inline void add (int from, int To, int Value)
{
to[++num]=To;
next[num]=head[from];
head[from]=num;
value[num]=Value;
}
inline int bfs(int x)
{
int Max = -0x7fffffff, p;
memset (done, 0, sizeof(done));
queue <int> q;
q.push (x);
done[x] = 1;
d[x] = 0;
int now;
while (!q.empty())
{
now = q.front();
if (d[now] > Max && now != x)
Max = d[now], p = now;
q.pop();
for (int i = head[now]; i; i=next[i])
if (!done[to[i]])
d[to[i]] = d[now] + value[i], done[to[i]] = 1, q.push (to[i]);
}
return p;
}
bool dfs (int x, int fa, int target)
{
if (x == target)
return 1;
for (int i = head[x]; i; i=next[i])
if (to[i] != fa)
{
if (dfs (to[i], x, target))
{
value[i] = value[i^1] = -1;
return 1;
}
}
return 0;
}
void dfs2 (int x, int fa)
{
int t1, t2;
t1 = t2 = 0;
f[x] = 0;
for (int i = head[x]; i; i = next[i])
if (to[i] != fa)
{
dfs2 (to[i], x);
if (f[to[i]] + value[i] > t1)
{
t2 = t1;
t1 = f[to[i]] + value[i];
}
else
if (f[to[i]] + value[i] > t2)
t2 = f[to[i]] + value[i];
}
f[x] = t1;
if (Max < t1 + t2)
Max = t1 + t2, tool = x;
}
int main()
{
scanf ("%d%d", &n, &k);
for (int i = 1, a1, a2; i <= n-1; i++)
scanf ("%d%d", &a1, &a2), add(a1, a2, 1), add(a2, a1, 1);
ans = 0;
int now = bfs (1);
int root = bfs (now);
ans = d[root] - 1;
dfs (now, 0, root);
if (k > 1)
{
dfs2 (1, -1);
ans += Max - 1;
}
printf ("%d\n", 2 * (n-1) - ans);
return 0;
}