洛谷P3018 [USACO11MAR]树装饰Tree Decoration
树形DP
因为要求最小,我们就贪心地用每个子树中的最小cost来支付就行了
1 #include <bits/stdc++.h> 2 #define For(i, j, k) for(int i=j; i<=k; i++) 3 #define Dow(i, j, k) for(int i=j; i>=k; i--) 4 #define LL long long 5 using namespace std; 6 inline int read() { 7 int x = 0, f = 1; 8 char ch = getchar(); 9 while(ch<‘0‘||ch>‘9‘) { if(ch==‘-‘) f = -1; ch = getchar(); } 10 while(ch>=‘0‘&&ch<=‘9‘) { x = x*10+ch-48; ch = getchar(); } 11 return x * f; 12 } 13 14 const int N = 1e5+11; 15 int n, nedge; 16 int Mn[N], fa[N], head[N], need[N]; 17 LL f[N], have[N]; 18 struct edge{ 19 int to, pre; 20 }e[N*2]; 21 22 inline void add(int x,int y) { 23 e[++nedge].to = y; 24 e[nedge].pre = head[x]; 25 head[x] = nedge; 26 } 27 28 void dfs(int u) { 29 for(int i=head[u]; i; i=e[i].pre) { 30 int v = e[i].to; 31 if(v == fa[u]) continue; 32 dfs( v ); 33 have[u] += have[v]; 34 Mn[u] = min(Mn[u], Mn[v]); 35 f[u] += f[v]; 36 } 37 if(need[u] > have[u]) { 38 f[u] = f[u]+1ll*(need[u]-have[u])*Mn[u]; 39 have[u] = need[u]; 40 } 41 } 42 43 int main() { 44 n = read(); 45 For(i, 1, n) { 46 fa[i] = read(); 47 need[i] = read(); 48 Mn[i] = read(); 49 if(fa[i] != -1) add(i, fa[i]), add(fa[i], i); 50 } 51 dfs(1); 52 printf("%lld\n",f[1]); 53 return 0; 54 }