BZOJ1912 最长链树形DP

每次求出最长链更新答案后要将最长链上的边权改为-1

写的贼长 还可以优化...

  1 /*Huyyt*/
  2 #include<bits/stdc++.h>
  3 #define mem(a,b) memset(a,b,sizeof(a))
  4 #define pb push_back
  5 using namespace std;
  6 typedef long long ll;
  7 typedef unsigned long long ull;
  8 using namespace std;
  9 const int MAXN = 1e5 + 5, MAXM = 2e5 + 5;
 10 int to[MAXM << 1], nxt[MAXM << 1], Head[MAXN], ed = 1;
 11 int value[MAXM << 1];
 12 inline void addedge(int u, int v, int val)
 13 {
 14         to[++ed] = v;
 15         nxt[ed] = Head[u];
 16         value[ed] = val;
 17         Head[u] = ed;
 18 }
 19 int d[MAXN];
 20 void dfs(int x, int pre)
 21 {
 22         for (int v, i = Head[x]; i; i = nxt[i])
 23         {
 24                 v = to[i];
 25                 if (v == pre)
 26                 {
 27                         continue;
 28                 }
 29                 d[v] = d[x] + value[i];
 30                 dfs(v, x);
 31         }
 32 }
 33 void change(int x)
 34 {
 35         for (int v, i = Head[x]; i; i = nxt[i])
 36         {
 37                 v = to[i];
 38                 if (d[v] == d[x] - 1)
 39                 {
 40                         value[i] = value[i ^ 1] = -1;
 41                         change(v);
 42                 }
 43         }
 44 }
 45 int s, t, dmx = -1;
 46 int ans2 = 0, vis[MAXN], dpd[MAXN];
 47 void dp(int x)
 48 {
 49         vis[x] = 1;
 50         for (int v, i = Head[x]; i; i = nxt[i])
 51         {
 52                 v = to[i];
 53                 if (vis[v])
 54                 {
 55                         continue;
 56                 }
 57                 dp(v);
 58                 ans2 = max(ans2, dpd[x] + dpd[v] + value[i]);
 59                 dpd[x] = max(dpd[x], dpd[v] + value[i]);
 60         }
 61 }
 62 int main()
 63 {
 64         int anser;
 65         int n, k;
 66         int u, v;
 67         scanf("%d %d", &n, &k);
 68         for (int i = 1; i < n; i++)
 69         {
 70                 scanf("%d %d", &u, &v);
 71                 addedge(u, v, 1), addedge(v, u, 1);
 72         }
 73         anser = 2 * (n - 1);
 74         d[1] = 0;
 75         dfs(1, 0);
 76         for (int i = 1; i <= n; i++)
 77         {
 78                 if (d[i] > dmx)
 79                 {
 80                         dmx = d[i];
 81                         s = i;
 82                 }
 83         }
 84         d[s] = 0;
 85         dfs(s, 0);
 86         dmx = -1;
 87         for (int i = 1; i <= n; i++)
 88         {
 89                 if (d[i] > dmx)
 90                 {
 91                         dmx = d[i];
 92                         t = i;
 93                 }
 94         }
 95         anser -= d[t] - 1;
 96         if (k == 1)
 97         {
 98                 printf("%d\n", anser);
 99                 return 0;
100         }
101         change(t);
102         dp(1);
103         anser -= ans2 - 1;
104         printf("%d\n", anser);
105         return 0;
106 }

//BZOJ1912