题解【洛谷P5765】[CQOI2005]珠宝

题面

考虑树形 DP。

设 \(dp_{i,j}\) 表示节点 \(i\) 染成颜色 \(j\)，\(i\) 子树内最小的编号总和。

容易得出转移方程：\(dp_{i,j}=j+\sum\limits_{v\in son_i, k\not =j}dp_{v,k}\)。

由四色定理可知，\(j\) 最大为 \(4\)，因此数组大小开到 \(5\) 即可。

#include <bits/stdc++.h>
#define DEBUG fprintf(stderr, "Passing [%s] line %d\n", __FUNCTION__, __LINE__)
#define File(x) freopen(x".in","r",stdin); freopen(x".out","w",stdout)
#define DC int T = gi <int> (); while (T--)

using namespace std;

typedef long long LL;
typedef pair <int, int> PII;
typedef pair <int, PII> PIII;

template <typename T>
inline T gi()
{
    T f = 1, x = 0; char c = getchar();
    while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
    while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return f * x;
}

const int INF = 0x3f3f3f3f, N = 50003, M = N << 1;

int n;
int tot, head[N], ver[M], nxt[M];
int dp[N][5];

inline void add(int u, int v) {ver[++tot] = v, nxt[tot] = head[u], head[u] = tot;}

void dfs(int u, int f)
{
	for (int i = 1; i <= 4; i+=1) dp[u][i] = i;
	for (int i = head[u]; i; i = nxt[i])
	{
		int v = ver[i];
		if (v == f) continue;
		dfs(v, u);
		for (int j = 1; j <= 4; j+=1)
		{
			int mn = INF;
			for (int k = 1; k <= 4; k+=1)
				if (j != k) mn = min(mn, dp[v][k]);
			dp[u][j] += mn;
		}
	}
}

int main()
{
    //File("");
    n = gi <int> ();
    for (int i = 1; i < n; i+=1)
    {
    	int u = gi <int> (), v = gi <int> ();
    	add(u, v), add(v, u);
    }
    dfs(1, 0);
    printf("%d\n", *min_element(dp[1] + 1, dp[1] + 1 + 4));
    return 0;
}