Given a graph, we define "proper coloring" as coloring of the graph nodes in such way that no two adjacent nodes have the same color. If we map each color to a positive integer, we can calculate the sum of all colors assigned to the graph.

In this problem you will be given a tree (connected graph with no simple loops). Can you determine what the minimum color sum can be achieved when the tree is properly colored? (Image to the right shows a proper coloring of the second example tree with sum=11)

Input

The input file consists of several test cases. Each test case starts with n (1 ≤ n ≤ 10000), the number of nodes in the tree. Next n lines will be of the form "u: v1 v2 ... vk" where u is the root of a subtree and vi‘s are its children (0 ≤ u, vi ≤ n-1).

Every test case will be followed by a blank line. Input ends with a case n=0, which should not be processed.

Output

For each test case print the minimum sum of colors that can be achieved by some proper coloring of the tree.

Sample Input

2
0:
1: 0

8
0: 1 2 3
1: 4 5
2:
3: 6 7
4:
5:
6:
7:

0

Output for the Sample Input

3
11

题意：给定一棵树，不同颜色用不同数字表示，相邻结点不能染相同颜色，问正确染完色后，整棵树的颜色加起来最小值是多少。

思路：树形DP，dp[u][i]代表结点u染颜色i的时候最小情况。然后就是常规的树形DP了。。

代码：

#include <stdio.h>
#include <string.h>
#include <vector>
#define INF 0x3f3f3f3f
#define min(a,b) ((a)<(b)?(a):(b))
using namespace std;

const int N = 10005;
int n, dp[N][10];
char str[25555];
vector<int> g[N];

void handle() {
	int i = 0;
	int len = strlen(str), num = -1;
	while (str[i] < ‘0‘ || str[i] > ‘9‘) i++;
	for (;str[i] != ‘:‘; i++) {
		if (num == -1)
			num = str[i] - ‘0‘;
		else
			num = num * 10 + str[i] - ‘0‘;
	}
	int u = num;
	num = -1;
	for (;i <= len; i++) {
		if (num != -1 && (str[i] < ‘0‘ || str[i] > ‘9‘)) {
			g[u].push_back(num);
			g[num].push_back(u);
			num = -1;
		}
		else if (str[i] >= ‘0‘ && str[i] <= ‘9‘) {
			if (num == -1)
				num = str[i] - ‘0‘;
			else
				num = num * 10 + str[i] - ‘0‘;
		}
	}
}

void dfs(int u, int fa) {
	int i;
	for (i = 1; i <= 6; i++)
		dp[u][i] = i;
	for (i = 0; i < g[u].size(); i++) {
		int v = g[u][i];
		if (v == fa) continue;
		dfs(v, u);
		for (int i = 1; i <= 6; i++) {
			int tmp = INF;
			for (int j = 1; j <= 6; j++) {
				if (i == j) continue;
				tmp = min(tmp, dp[v][j]);
			}
			dp[u][i] += tmp;
		}
	}
}

int main() {
    while (~scanf("%d%*c", &n) && n) {
		memset(g, 0, sizeof(g));
		while (gets(str) && str[0] != ‘\0‘) {
			handle();
		}
		dfs(0, -1);
		int ans = INF;
		for (int i = 1; i <= 6; i++)
			ans = min(ans, dp[0][i]);
		printf("%d\n", ans);
	}
    return 0;
}

UVA 11307 - Alternative Arborescence（dp),布布扣,bubuko.com

UVA 11307 - Alternative Arborescence（dp)

原文：http://blog.csdn.net/accelerator_/article/details/22165353