题面
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.
题解
设f[i][j]表示从i开始，j步以内的牛的数量
很容易想到f[i][j]=sum(f[k][j-1])再去减去一堆什么东西
（k表示和i相连的节点）
我这个蒟蒻尽然用容斥原理做。。。。
要不是题目中的K很小，我觉得会TLE。。。
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
#define MAX 100100
inline int read()
{
register int x=0,t=1;
register char ch=getchar();
while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
if(ch=='-'){t=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-48;ch=getchar();}
return x*t;
}
struct Line
{
int u,v;
}e[MAX];
long long N,K,f[MAX][30],Ans[MAX];
int main()
{
N=read();K=read();
register int u,v;
for(int i=1;i<N;++i)
e[i]=(Line){read(),read()};
for(int i=1;i<=N;++i)
f[i][0]=read();
//f[i][j]表示从i节点开始走j步的奶牛数
for(int i=1;i<N;++i)
{
f[e[i].u][1]+=f[e[i].v][0];
f[e[i].v][1]+=f[e[i].u][0];
}
for(int i=2;i<=K;++i)
{
for(int j=1;j<N;++j)//枚举边
{
for(int k=i-1,t=1;k>=0;t=!t,k--)//容斥大法
{
if(t)
{
f[e[j].u][i]+=f[e[j].v][k];
f[e[j].v][i]+=f[e[j].u][k];
}
else
{
f[e[j].u][i]-=f[e[j].u][k];
f[e[j].v][i]-=f[e[j].v][k];
}
}
}
}
for(int i=1;i<=N;++i)
for(int j=0;j<=K;++j)
Ans[i]+=f[i][j];
for(int i=1;i<=N;++i)
printf("%d\n",Ans[i]);
return 0;
}