codeforces 416B. Appleman and Tree 树形dp

题目链接

 

Fill a DP table such as the following bottom-up:

• DP[v][0] = the number of ways that the subtree rooted at vertex v has no black vertex.
• DP[v][1] = the number of ways that the subtree rooted at vertex v has one black vertex.

The recursion pseudo code is folloing:

DFS(v): DP[v][0] = 1 DP[v][1] = 0 foreach u : the children of vertex v DFS(u) DP[v][1] *= DP[u][0] DP[v][1] += DP[v][0]*DP[u][1] DP[v][0] *= DP[u][0] if x[v] == 1: DP[v][1] = DP[v][0] else: DP[v][0] += DP[v][1]

The answer is DP[root][1]

 

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
const int maxn = 1e5+2;
ll dp[maxn][2];
int head[maxn], num, k, a[maxn];
struct node
{
    int to, nextt;
}e[maxn*2];
void add(int u, int v) {
    e[num].to = v, e[num].nextt = head[u], head[u] = num++;
}
void init() {
    num = 0;
    mem1(head);
}
void dfs(int u, int fa) {
    dp[u][1] = 0;
    dp[u][0] = 1;
    for(int i = head[u]; ~i; i = e[i].nextt) {
        int v = e[i].to;
        if(v == fa)
            continue;
        dfs(v, u);
        dp[u][1] = dp[u][1]*dp[v][0]%mod;
        dp[u][1] = (dp[u][1]+dp[u][0]*dp[v][1]%mod)%mod;
        dp[u][0] = dp[u][0]*dp[v][0]%mod;
    }
    if(a[u]) {
        dp[u][1] = dp[u][0];
    } else {
        dp[u][0] = (dp[u][0]+dp[u][1])%mod;
    }
}
int main()
{
    int n, x, y;
    cin>>n;
    init();
    for(int i = 1; i<n; i++) {
        scanf("%d", &x);
        add(x, i);
        add(i, x);
    }
    for(int i = 0; i<n; i++)
        scanf("%d", &a[i]);
    dfs(0, -1);
    cout<<dp[0][1]<<endl;
    return 0;
}

 

转载于:https://www.cnblogs.com/yohaha/p/5262846.html