[bzoj 1060][luogu p1131]时态同步

Posted 2020-10-09

(似乎是)第一次做树形DP吔,好鸡冻~

 

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题目大意:

一棵有根边权树,修改边权值代价为边权值变化量,目标状态是每个叶子到根的路径权值相等.求达到目标状态耗费的最小代价.

 

要把每条路径的权值加到最大权值,可以证明它最优然而我不会,递归处理子树即可.

 

代码如下:

 

 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 }

 

用了读优之后神清气爽,

BZOJ上432ms Rk15,luogu上48ms Rk 3.

 

(辣鸡B(婊)Z(子)O(口)J(丁))