POJ 1849. Two 树形DP解法 c++代码

题目等价于：

所有边权值和的两倍 - 树直径上边的权值和

解题思路：

树形dp

dp[i][0]：结点i最长枝的长度

dp[i][1]：结点i第二长枝的长度

对所有结点的dp[i][0] + dp[i][1]求最大

代码：

#include 
#include 
using namespace std;
#define maxN 100005

int head[maxN], dp[maxN][2];
bool visited[maxN];
int cnt = 0, dmtr = 0;

struct Edge {
	int v, w, next;
} edges[maxN - 1];

void addEdge(int u, int v, int w) {
	edges[cnt].v = v;
	edges[cnt].w = w;
	edges[cnt].next = head[u];
	head[u] = cnt++;
}

void treeDp(int root) {
	visited[root] = true;
	int fstMax = 0, sndMax = 0;
	for (int i = head[root]; i != -1; i = edges[i].next) {
		int v = edges[i].v;
		if (!visited[v]) {
			treeDp(v);
			int curr = dp[v][0] + edges[i].w; 
			if (curr > fstMax) {
				sndMax = fstMax;
				fstMax = curr;
			} else if (curr > sndMax) {
				sndMax = curr;
			}
		}
	}
	dp[root][0] = fstMax;
	dp[root][1] = sndMax;
	if (fstMax + sndMax > dmtr) dmtr = fstMax + sndMax;
}

int main(void) {
	int N, start;
	int u, v, w;
	int circum = 0;
	memset(head, -1, sizeof(head));
	memset(visited, 0, sizeof(visited));
	memset(dp, 0, sizeof(dp));
	scanf("%d%d", &N, &start);
	for (int i = 0; i < N - 1; i++) {
		scanf("%d%d%d", &u, &v, &w);
		circum += w;
		addEdge(u, v, w);
		addEdge(v, u, w);
	}
	circum *= 2;
	treeDp(start);
	printf("%d\n", circum - dmtr);
}