題解【洛谷P1352】沒有上司的舞會

題面c++

題解

樹形\(\text{DP}\)入門題。spa

咱們設\(dp[i][0/1]\)表示第\(i\)個節點選\(/\)不選的最大快樂指數。code

狀態轉移方程:
\(dp[i][0]=a[i]+\sum_{v∈son[u]}dp[v][1]\),其中\(a[i]\)爲每一個員工的快樂指數。
\(dp[i][1]=\sum_{v∈son[u]}\max{(dp[v][1],dp[v][0])}\)get

答案爲\(\max{(dp[rt][0],dp[rt][1])}\),其中\(rt\)爲沒有上司的員工。it

轉移一下便可。入門

代碼

#include <bits/stdc++.h>
#define itn int
#define gI gi

using namespace std;

inline int gi()
{
    int f = 1, x = 0; char c = getchar();
    while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
    while (c >= '0' && c <= '9') x = x * 10 + (c ^ 48), c = getchar();
    return f * x;
}

const int maxn = 6003;

int n, a[maxn], tot, head[maxn], ver[maxn * 2], nxt[maxn], ans, vis[maxn], rt;
int dp[maxn][2];//0:xuan 1:buxuan

inline void add(int u, int v)
{
    ver[++tot] = v, nxt[tot] = head[u], head[u] = tot;
}

void dfs(int u, int f)
{
    dp[u][0] = a[u];
    for (int i = head[u]; i; i = nxt[i])
    {
        int v = ver[i];
        if (v == f) continue;
        dfs(v, u);
        dp[u][0] += dp[v][1];
        dp[u][1] += max(dp[v][0], dp[v][1]);//狀態轉移
    }
}

int main()
{
    n = gi();
    for (int i = 1; i <= n; i+=1) a[i] = gi();
    for (int i = 1; i < n; i+=1)
    {
        int u = gi(), v = gi();
        add(u, v);
        add(v, u);
        vis[u] = 1;
    }
    int h = gi(), o = gi();
    for (int i = 1; i <= n; i+=1) if (!vis[i]) {rt = i; break;}//找到根節點,即沒有上司的員工編號
    dfs(rt, 0);
    printf("%d\n", max(dp[rt][0], dp[rt][1]));//答案就是根節點選/不選取max
    return 0;
}

總結

由此,咱們能夠得出樹形\(\text{DP}\)的狀態的基本形式:class

\(dp[i][…]\)表示第\(i\)個節點的狀態。總結

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