(似乎是)第一次做樹形DP吔,好雞凍~

題目大意:

一棵有根邊權樹,修改邊權值代價為邊權值變化量,目標狀態是每個葉子到根的路徑權值相等.求達到目標狀態耗費的最小代價.

要把每條路徑的權值加到最大權值,可以證明它最優然而我不會,遞歸處理子樹即可.

代碼如下:

 1 #include <cstdio>
 2 #include <cstring>
 3 #define max(a,b) (((a) > (b)) ? (a) : (b))
 4 #define mcl(a) memset(a,0,sizeof(a))
 5 const int MAXS = 60
 
 * 1024 * 1024,N = 500005;
 6 inline char __getchar()
 7 {
 8     static char buf[MAXS];static int len = 0,pos = 0;
 9     if(pos == len)pos = 0,len = fread(buf,1,MAXS,stdin);if(pos == len)return ‘\0‘;return buf[pos++];
10 }
11 inline int getint()
12 {
13     int register num = 0;char register ch = __getchar();for
 
(;ch < ‘0‘ || ch > ‘9‘;ch = __getchar());
14     for(;ch >= ‘0‘ && ch <= ‘9‘;ch = __getchar())num = (num << 3) + (num << 1) + (ch ^ ‘0‘);return num;
15 }
16 int n,s,cnt = 0,head[N];long long f[N],ans;
17 struct edge{int next,to,val;}e[N << 1];
18 void addedge(int u,int
 
 v,int w){e[++cnt].to = v;e[cnt].next = head[u];head[u] = cnt;e[cnt].val = w;}
19 void dfs(int u,int fa)
20 {
21     for(int register i = head[u];i;i = e[i].next)
22     {
23         int register v = e[i].to;if(v == fa)continue;
24         dfs(v,u);
25         f[u] = max(f[u],f[v] + e[i].val);
26     }
27     for(int register i = head[u];i;i = e[i].next) if(fa != e[i].to) ans += f[u] - f[e[i].to] - e[i].val;
28 }
29 int main()
30 {
31     mcl(head);mcl(e);mcl(f);
32     n = getint();s = getint();
33     for(int register i = 1;i < n;i++){int register a,b,t;a = getint();b = getint();t = getint();addedge(a,b,t);addedge(b,a,t);}
34     dfs(s,0);printf("%lld\n",ans);
35     return 0;
36 }

用了讀優之後神清氣爽,

BZOJ上432ms Rk15,luogu上48ms Rk 3.

(辣雞B(婊)Z(子)O(口)J(丁))

[bzoj 1060][luogu p1131]時態同步