洛谷P3018 [USACO11MAR]树装饰Tree Decoration

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

#include <bits/stdc++.h>
#define For(i, j, k) for(int i=j; i<=k; i++)
#define Dow(i, j, k) for(int i=j; i>=k; i--)
#define LL long long
using namespace std;
inline int read() {
    int x = 0, f = 1;
    char ch = getchar();
    while(ch<'0'||ch>'9') { if(ch=='-') f = -1; ch = getchar(); }
    while(ch>='0'&&ch<='9') { x = x*10+ch-48; ch = getchar(); }
    return x * f;
}

const int N = 1e5+11; 
int n, nedge; 
int Mn[N], fa[N], head[N], need[N]; 
LL f[N], have[N];  
struct edge{
    int to, pre; 
}e[N*2];

inline void add(int x,int y) {
    e[++nedge].to = y; 
    e[nedge].pre = head[x]; 
    head[x] = nedge; 
}

void dfs(int u) {
    for(int i=head[u]; i; i=e[i].pre) {
        int v = e[i].to; 
        if(v == fa[u]) continue; 
        dfs( v ); 
        have[u] += have[v]; 
        Mn[u] = min(Mn[u], Mn[v]); 
        f[u] += f[v]; 
    }
    if(need[u] > have[u]) {
        f[u] = f[u]+1ll*(need[u]-have[u])*Mn[u]; 
        have[u] = need[u]; 
    }
}

int main() {
    n = read(); 
    For(i, 1, n) {
        fa[i] = read(); 
        need[i] = read(); 
        Mn[i] = read(); 
        if(fa[i] != -1) add(i, fa[i]), add(fa[i], i); 
    }
    dfs(1); 
    printf("%lld
",f[1]); 
    return 0; 
}

转载于:https://www.cnblogs.com/third2333/p/8444558.html