P3047 [USACO12FEB]附近的牛Nearby Cows

題目描述

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.

給出一棵n個點的樹，每個點上有C_i頭牛，問每個點k步範圍內各有多少頭牛。

輸入輸出格式

• 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)

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

輸入輸出樣例

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

15 
21 
16 
10 
8 
11 

說明

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.

思路：樹形dp+容斥原理。

錯因：數組開小了。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define MAXN 100010
using namespace std;
int n,y,tot;
int val[MAXN],into[MAXN],f[MAXN][21];
int to[MAXN*2],net[MAXN*2],head[MAXN*2];
void add(int u,int v){
    to[++tot]=v;net[tot]=head[u];head[u]=tot;
    to[++tot]=u;net[tot]=head[v];head[v]=tot;
}
int main(){
    //freopen("young.in","r",stdin);
    //freopen("young.out","w",stdout);
    scanf("%d%d",&n,&y);
    for(int i=1;i<n;i++){
        int u,v;
        scanf("%d%d",&u,&v);
        add(u,v);
        into[u]++;
        into[v]++;
    }
    for(int i=1;i<=n;i++)    scanf("%d",&val[i]);
    for(int i=1;i<=n;i++)    f[i][0]=val[i];
    for(int j=1;j<=y;j++)
        for(int i=1;i<=n;i++){
            for(int k=head[i];k;k=net[k])
                f[i][j]+=f[to[k]][j-1];
            if(j>1)    f[i][j]-=(into[i]-1)*f[i][j-2];
            else f[i][1]+=f[i][0];
        }
    for(int i=1;i<=n;i++)
        cout<<f[i][y]<<endl;
}

洛谷 P3047 [USACO12FEB]附近的牛Nearby Cows