标签:代码 algorithm space pen temp char iostream 子节点 sdi
题意让我们用最少的代价把叶子节点到根节点的距离调成相同
每次向最长链看齐即可
令$f[u]$表示以$u$为根到叶节点的最长链长度
则$$f[u] = max\left\{f[v]+e(u,v)\right\}$$
统计答案
$$ans = \sum _{v(e(u,v)\in E)} f[u]-(f[v]+e(u,v))$$
上代码
1 #include <iostream> 2 #include <algorithm> 3 #include <cstdio> 4 #include <cstring> 5 #define ll long long 6 using namespace std; 7 8 template <typename T> void in(T &x) { 9 x = 0; T f = 1; char ch = getchar(); 10 while(!isdigit(ch)) {if(ch == ‘-‘) f = -1; ch = getchar();} 11 while( isdigit(ch)) {x = 10 * x + ch - 48; ch = getchar();} 12 x *= f; 13 } 14 15 template <typename T> void out(T x) { 16 if(x < 0) x = -x , putchar(‘-‘); 17 if(x > 9) out(x/10); 18 putchar(x%10 + 48); 19 } 20 //------------------------------------------------------- 21 22 const int N = 5e5+7; 23 24 int n,s; 25 ll ans,f[N]; 26 27 struct edge { 28 int v,w,nxt; 29 edge(int v = 0,int w = 0,int nxt = 0):v(v),w(w),nxt(nxt){}; 30 }e[N<<1]; int head[N],e_cnt; 31 32 void add(int u,int v,int w) { 33 e[++e_cnt] = edge(v,w,head[u]); head[u] = e_cnt; 34 } 35 36 void dfs(int u,int fa) { 37 int i; 38 int max_e = 0; 39 for(i = head[u]; i;i = e[i].nxt) { 40 int v = e[i].v; if(v == fa) continue; 41 dfs(v,u); 42 f[u] = max(f[u],f[v]+e[i].w); 43 } 44 for(i = head[u]; i;i = e[i].nxt) { 45 int v = e[i].v; if(v == fa) continue; 46 ans += f[u] - (f[v]+e[i].w); 47 } 48 } 49 50 int main() { 51 int i,u,v,w; 52 in(n); in(s); 53 for(i = 1;i < n; ++i) { 54 in(u); in(v); in(w); 55 add(u,v,w); add(v,u,w); 56 } 57 dfs(s,0); 58 out(ans); 59 return 0; 60 }
标签:代码 algorithm space pen temp char iostream 子节点 sdi
原文地址:https://www.cnblogs.com/mzg1805/p/11355790.html