POJ 1849 Two (树形dp 树的直径 两种方法)

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
int const MAX = 100005;
int n, s, cnt;
int dis[MAX], head[MAX];

struct EGDE
{
    int v, w, next; 
}e[MAX];

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

void DFS(int u, int fa, int d)
{
    for(int i = head[u]; i != -1; i = e[i].next)
    {
        int v = e[i].v;
        int w = e[i].w;
        if(v != fa)
        {
            dis[v] = w + d;
            DFS(v, u, w + d);
        }
    }
    return;
}

int main()
{
    cnt = 0;
    int sum = 0;
    scanf("%d %d", &n, &s);
    memset(head, -1, sizeof(head));
    memset(dis, 0, sizeof(dis));
    for(int i = 0; i < n - 1; i++)
    {
        int u, v, w;
        scanf("%d %d %d", &u, &v, &w);
        Add(u, v, w);
        Add(v, u, w);
        sum += 2 * w;
    }
    dis[s] = 0;
    DFS(s, -1, 0);
    int o, ma = -1;
    for(int i = 1; i <= n; i++)
    {
        if(dis[i] > ma)
        {
            ma = dis[i];
            o = i;
        }
    }
    dis[o] = 0;
    DFS(o, -1, 0);
    ma = -1;
    for(int i = 1; i <= n; i++)
        if(dis[i] > ma)
            ma = dis[i];
    printf("%d\n", sum - ma);
}

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int const MAX = 100005;
int head[MAX], dp[MAX][2];
int n, s, cnt, ans;

struct EDGE
{
    int v, w, next;
}e[MAX];

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

void DFS(int u, int fa)
{
    dp[u][0] = dp[u][1] = 0;
    for(int i = head[u]; i != -1; i = e[i].next)
    {
        int v = e[i].v;
        int w = e[i].w;
        if(v != fa)
        {
            DFS(v, u);
            if(dp[u][0] < dp[v][0] + w)
            {
                int tmp = dp[u][0];
                dp[u][0] = dp[v][0] + w;
                dp[u][1] = tmp;
            }
            else if(dp[u][1] < dp[v][0] + w)
                dp[u][1] = dp[v][0] + w;
        }
    }
    ans = max(ans, dp[u][1] + dp[u][0]);
    return;
}

int main()
{
    cnt = 0;
    ans = 0;
    memset(head, -1, sizeof(head));
    scanf("%d %d", &n, &s);
    int sum = 0;
    for(int i = 0; i < n - 1; i++)
    {
        int u, v, w;
        scanf("%d %d %d", &u, &v, &w);
        Add(u, v, w);
        Add(v, u, w);
        sum += 2 * w;
    }
    DFS(s, -1);
    printf("%d\n", sum - ans);
}