Codeforces Round #592 (Div. 2)

D. Paint the Tree

题意：
给出一棵树，有三种颜色，要求任意链上的三点的颜色是不一样的，每个点图3种颜色都有自己的花费值，问最小的花费
Solution:
显然一个点度超过2，那么这3个点颜色都不同，无解
所以度都为1，2。树退化成链。
确定了第1,2个，第3个点也唯一确定了（不和1,2的一样），那第4个也出来了
所以两个点就能确定所有点。枚举前两个点颜色就行

/**/
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cctype>
#include <iostream>
#include <algorithm>
#include <map>
#include <set>
#include <vector>
#include <string>
#include <stack>
#include <queue>

typedef long long LL;
typedef unsigned long long ULL;
using namespace std;

bool Sqrt(LL n) { return (LL)sqrt(n) * sqrt(n) == n; }
const double PI = acos(-1.0), ESP = 1e-10;
const LL INF = 99999999999999;
const int inf = 999999999, N = 1e5 + 5;
int n, u, v, rt, a[] = {1, 2, 3}, ans[N], cur[N], c[4][N], deg[N];
LL res = 1e18, cost;
vector<int> G[N];

void dfs(int u, int h, int p = 0)
{
	cost += c[a[h]][u];
	cur[u] = h;
	for(auto v : G[u]) if(v != p)
		dfs(v, (h + 1) % 3, u);
}

void solve()
{
	cost = 0;
	dfs(rt, 0);
	if(cost < res) {
		res = cost;
		for(int i = 1; i <= n; i++) ans[i] = a[cur[i]];
	}
}

int main()
{
	//freopen("in.txt", "r", stdin);
	//freopen("out.txt", "w", stdout);
	scanf("%d", &n);
	for(int i = 1; i <= 3; i++) for(int j = 1; j <= n; j++) scanf("%d", &c[i][j]);
	for(int i = 1; i < n; i++) {
		scanf("%d%d", &u, &v);
		G[u].push_back(v); G[v].push_back(u);
		deg[v]++; deg[u]++;
	}
	rt = -1;
	for(int i = 1; i <= n; i++) {
		if(deg[i] > 2) { puts("-1"); exit(0);}
		if(deg[i] == 1 && rt == -1) rt=  i;
	}
	do {
		solve();
	}while(next_permutation(a, a + 3));
	printf("%lld
", res);
	for(int i = 1; i < n; i++) printf("%d ", ans[i]);
	printf("%d
", ans[n]);

	return 0;
}
/*
    input:
    output:
    modeling:
    methods:
    complexity:
    summary:
*/

C. The Football Season
简单数学，打的时候没有发现限制条件x < w，T了；后来狂想exgcd，emmmmm

LL n, p, w, d;

int main()
{
	//freopen("in.txt", "r", stdin);
	//freopen("out.txt", "w", stdout);
	while(scanf("%lld%lld%lld%lld", &n, &p, &w, &d) == 4) {
		bool flag = 1;
		for(LL x = 0, y; x < w; x++) {
			LL k = p - d * x;
			if(k % w == 0) { 
				y = k / w;
				if(y >= 0 && x + y <= n) { flag = 0; printf("%lld %lld %lld
", y, x, n - x - y); break;}
			}
		}
		if(flag) puts("-1");
	}

	return 0;
}