Paint the Tree【Codeforces Round #592 (Div. 2) D】【枚举+dfs】

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Codeforces 1244 D Codeforces Round #592 (Div. 2)

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  这道题要问的是对于给出一棵树，对于任意链上的三点的颜色是不一样的，三个都要求不同，然后问最小的花费是什么？

  我们看到，假如有个点的度是大于等于3的时候，也就是它链接了3个以上的点的话，一定是不可能满足条件的。

  那么不就是从一个度为1的点出发，然后呢，我们枚举前两个点的颜色，一共有6种可能，然后呢，就是去往下跑即可。

//#pragma comment(linker, "/STACK:102400000,102400000")
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define MIN3(a, b, c) min(min(a, b), c)
#define MAX3(a, b, c) max(a, max(b, c))
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 1e5 + 7;
int N, head[maxN], cnt, ans[maxN], cop[maxN], du[maxN] = {0};
ll cost[maxN][4];
inline int diff(int e1, int e2)
{
    if(e1 == 1 && e2 == 2) return 3;
    else if(e1 == 1 && e2 == 3) return 2;
    else if(e1 == 2 && e2 == 1) return 3;
    else if(e1 == 2 && e2 == 3) return 1;
    else if(e1 == 3 && e2 == 1) return 2;
    else return 1;
}
struct Eddge
{
    int nex, to;
    Eddge(int a=-1, int b=0):nex(a), to(b) {}
}edge[maxN<<1];
inline void addEddge(int u, int v)
{
    edge[cnt] = Eddge(head[u], v);
    head[u] = cnt++;
}
inline void _add(int u, int v) { addEddge(u, v); addEddge(v, u); }
ll minn_cost = INF, now_cost;
void dfs(int u, int col_1, int col_2, int fa)
{
    int now_col = diff(col_1, col_2);
    cop[u] = now_col;
    now_cost += cost[u][now_col];
    for(int i=head[u], v; ~i; i=edge[i].nex)
    {
        v = edge[i].to;
        if(v == fa) continue;
        dfs(v, col_2, now_col, u);
    }
}
inline void init()
{
    cnt = 0;
    for(int i=1; i<=N; i++) head[i] = -1;
}
int main()
{
    scanf("%d", &N);
    init();
    for(int col=1; col<=3; col++)
    {
        for(int i=1; i<=N; i++)
        {
            scanf("%lld", &cost[i][col]);
        }
    }
    for(int i=1, u, v; i<N; i++)
    {
        scanf("%d%d", &u, &v);
        _add(u, v);
        du[u]++; du[v]++;
    }
    int rt = 0;
    for(int i=1; i<=N; i++) if(du[i] >= 3) { printf("-1\n"); return 0; }
    for(int i=1; i<=N; i++) if(du[i] == 1) { rt = i; break; }
    for(int col=1; col<=3; col++)
    {
        cop[rt] = col;
        for(int colful=1, u; colful<=3; colful++)
        {
            if(colful == col) continue;
            now_cost = cost[rt][col];
            u = edge[head[rt]].to;
            cop[u] = colful;
            now_cost += cost[u][colful];
            for(int i=head[u], v; ~i; i=edge[i].nex)
            {
                v = edge[i].to;
                if(v == rt) continue;
                dfs(v, col, colful, u);
            }
            if(minn_cost > now_cost)
            {
                minn_cost = now_cost;
                for(int i=1; i<=N; i++) ans[i] = cop[i];
            }
        }
    }
    printf("%lld\n", minn_cost);
    for(int i=1; i<=N; i++) printf("%d%c", ans[i], i == N ? '\n' : ' ');
    return 0;
}