历届试题 大臣的旅费 树形DP

题目链接：大臣的旅费

思路：锦囊说用广搜，可惜这题没说数据范围，担心复杂度太高，我就直接用的树形DP--求树的最远路径。

以城市1为整棵树的根结点，d(i)表示以i为根结点的子树的最远路径，还有一个f(i)表示以i为根结点，从根节点i出发的最远距离，状态转移方程d(i) = max1 + max2，或则d(i) = max，即选取最长的两条子树的最远路径相加，或只有一个子节点就是加上最大值。f(i) = max(len(i, j) + f(j));

AC代码：

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <utility>
#include <string>
#include <iostream>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <stack>
using namespace std;
#define eps 1e-10
#define inf 0x3f3f3f3f
#define PI pair<int, int>
const int maxn = 1e4 + 5;

vector<PI>son[maxn];
int d[maxn];
bool cmp(int a, int b) {
	return a > b;
}

int dfs(int u, int pre) {
	int n = son[u].size();
	d[u] = 0;
	int ans = 0;
	vector<int>road;
	if(!n) return 0;
	for(int i = 0; i < n; ++i) {
		int v = son[u][i].first, cost = son[u][i].second;
		if(v == pre) continue;
		int h = dfs(v, u);
		ans =max(ans, h + cost);
		road.push_back(h + cost);
	}
	sort(road.begin(), road.end(), cmp);
	if(road.size() == 1) d[u] = road[0];
	else if(road.size() > 1)d[u] = road[0] + road[1];
	return ans;
}

int main() {
	int n;
	while(scanf("%d", &n) == 1) {
		memset(d, 0, sizeof(d));
		for(int i = 0; i < n; ++i) son[i].clear();
		int x, y, cost;
		for(int i = 1; i < n; ++i) {
			scanf("%d%d%d", &x, &y, &cost);
			son[x-1].push_back(make_pair(y-1, cost));
			son[y-1].push_back(make_pair(x-1, cost));
		}
		dfs(0, -1);
		int ans = 0;
		for(int i = 0; i < n; ++i) {
			ans = max(ans, d[i]);
		}
		printf("%d\n", (ans + 1)*ans/2 + ans * 10);
	}
	return 0;
}

如有不当之处欢迎指出！