洛谷P3047 [USACO12FEB]Nearby Cows(树形dp)

P3047 [USACO12FEB]附近的牛Nearby Cows

题目描述

Farmer John has noticed that his cows often move between nearby fields. Taking this into account, he wants to plant enough grass in each of his fields not only for the cows situated initially in that field, but also for cows visiting from nearby fields.

Specifically, FJ's farm consists of N fields (1 <= N <= 100,000), where some pairs of fields are connected with bi-directional trails (N-1 of them in total). FJ has designed the farm so that between any two fields i and j, there is a unique path made up of trails connecting between i and j. Field i is home to C(i) cows, although cows sometimes move to a different field by crossing up to K trails (1 <= K <= 20).

FJ wants to plant enough grass in each field i to feed the maximum number of cows, M(i), that could possibly end up in that field -- that is, the number of cows that can potentially reach field i by following at most K trails. Given the structure of FJ's farm and the value of C(i) for each field i, please help FJ compute M(i) for every field i.

给出一棵n个点的树，每个点上有C_i头牛，问每个点k步范围内各有多少头牛。

输入输出格式

输入格式：

• Line 1: Two space-separated integers, N and K.

• Lines 2..N: Each line contains two space-separated integers, i and j (1 <= i,j <= N) indicating that fields i and j are directly connected by a trail.

• Lines N+1..2N: Line N+i contains the integer C(i). (0 <= C(i) <= 1000)

输出格式：

• Lines 1..N: Line i should contain the value of M(i).

输入输出样例

输入样例#1：

6 2
5 1
3 6
2 4
2 1
3 2
1
2
3
4
5
6

输出样例#1：

15
21
16
10
8
11

说明

There are 6 fields, with trails connecting (5,1), (3,6), (2,4), (2,1), and (3,2). Field i has C(i) = i cows.

Field 1 has M(1) = 15 cows within a distance of 2 trails, etc.

/*
树形dp:dp[i][j]:编号为i的节点向子节点走0~j步总和
预处理dp[i][j]数组,从当前点向父亲节点转移
容斥原理:ans+=dp[now][k],ans+=dp[father][k-1],ans-=dp[now][k-2] 不懂可手动模拟
*/
#include<iostream>
#include<cstdio>
#include<cstring>

#define N 100007
#define M 27

using namespace std;
int w[N],f[N],head[N],dp[N][M];
int n,m,k,cnt;
struct edge
{
    int u,to,pre;
}e[N<<];

inline int init()
{
    int x=,f=;char c=getchar();
    while(c>''||c<''){if(c=='-')f=-;c=getchar();}
    while(c>=''&&c<=''){x=x*+c-'';c=getchar();}
    return x*f;
}

inline void add(int u,int to)
{
    e[++cnt].to=to;e[cnt].pre=head[u];head[u]=cnt;
} 

void dfs(int from,int now)
{
    f[now]=from;dp[now][]=w[now];
    for(int i=head[now];i;i=e[i].pre)
    {
        if(e[i].to!=from)
        {
            dfs(now,e[i].to);
            for(int j=;j<=k;j++)
            dp[now][j]+=dp[e[i].to][j-];
        }
    }
}

void DP(int now)
{
    int K=k,ans=;ans=dp[now][k];
    while(now!= && K)
    {
        K--;ans+=dp[f[now]][K];
        if(K) ans-=dp[now][K-];
        now=f[now];
    }
    printf("%d\n",ans);
}

int main()
{
    int x,y;
    n=init();k=init();
    for(int i=;i<n;i++)
    {
        x=init();y=init();
        add(x,y);add(y,x);
    }
    for(int i=;i<=n;i++) w[i]=init();
    dfs(-,);
    for(int i=;i<=n;i++) for(int j=;j<=k;j++)//dfs时dp[i][j]是第i点刚好走j步，现在求前缀和
      dp[i][j]+=dp[i][j-];
    for(int i=;i<=n;i++) DP(i);
    return ;
}