題目鏈接

BZOJ4557

題解

orz
比較難的樹形dp
不過想想也還好

看數據猜狀態，一維是點，一維是D
那麽就先設\(f[i][j]\)表示\(i\)所在子樹已處理完畢，還能向上【或向任意方向】覆蓋\(j\)層的最小代價
考慮轉移，會發現子樹間會相互影響，一個子樹用\(f[s][j + 1]\)更新了\(f[i][j]\)，其它的子樹就完全沒必要再用\(f[s‘][j + 1]\)去更新了，此時反而可以用\(f[i][j]\)來減少該子樹付出的代價
所以我們設一個\(g[i][j]\)表示\(i\)為根的子樹前\(j\)層待覆蓋，\(j\)層以下已處理完畢的最小代價

\(f[i][j]\)

狀態轉移：枚舉子樹\(s\)
\[f[i][j] = min\{f[i][j] + g[s][j],f[s][j + 1] + g[i][j + 1]\}\]
\[g[i][j] += g[s][j - 1]\]

初始化就考慮\(i\)號點是不是一定被覆蓋，就可以確定\(f[i][0]\)和\(g[i][0]\)的初值

#include<iostream>
#include<cstdio>
#include<cmath>
 

#include<cstring>
#include<algorithm>
#define LL long long int
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define BUG(s,n) for (int i = 1; i <= (n); i++) cout<<s[i]<<‘ ‘; puts("");
using namespace std;
const
 
 int maxn = 500005,INF = 1000000000;
inline int read(){
    int out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if (c == ‘-‘) flag = -1; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    return out * flag;
}
int h[maxn],ne = 2;
struct EDGE{int to,nxt;}ed[maxn << 1];
inline void build(int u,int v){
    ed[ne] = (EDGE){v,h[u]}; h[u] = ne++;
    ed[ne] = (EDGE){u,h[v]}; h[v] = ne++;
}
int n,m,D,f[maxn][23],g[maxn][23],val[maxn],dan[maxn],fa[maxn];
int son[maxn],si;
void dfs(int u){
    for (int i = 1; i <= D; i++) f[u][i] = val[u];
    if (dan[u]) f[u][0] = g[u][0] = val[u];
    f[u][D + 1] = INF;
    Redge(u) if ((to = ed[k].to) != fa[u]){
        fa[to] = u; dfs(to);
        for (int i = 0; i <= D; i++)
            f[u][i] = min(f[u][i] + g[to][i],f[to][i + 1] + g[u][i + 1]);
        for (int i = D; i >= 0; i--)
            f[u][i] = min(f[u][i],f[u][i + 1]);
        g[u][0] = f[u][0];
        for (int i = 1; i <= D; i++)
            g[u][i] += g[to][i - 1];
        for (int i = 1; i <= D; i++)
            g[u][i] = min(g[u][i],g[u][i - 1]);
    }
}
int main(){
    n = read(); D = read();
    REP(i,n) val[i] = read();
    m = read();
    REP(i,m) dan[read()] = true;
    for (int i = 1; i < n; i++) build(read(),read());
    dfs(1);
    printf("%d\n",f[1][0]);
    return 0;
}

BZOJ4557 [JLoi2016]偵察守衛 【樹形dp】