POJ 1849 Two (树形dp 树的直径 两种方法)

Two

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 1232 Accepted: 619

Description

The city consists of intersections and streets that connect them.

Heavy snow covered the city so the mayor Milan gave to the winter-service a list of streets that have to be cleaned of snow. These streets are chosen such that the number of streets is as small as possible but still every two intersections to be connected i.e. between every two intersections there will be exactly one path. The winter service consists of two snow plovers and two drivers, Mirko and Slavko, and their starting position is on one of the intersections.

The snow plover burns one liter of fuel per meter (even if it is driving through a street that has already been cleared of snow) and it has to clean all streets from the list in such order so the total fuel spent is minimal. When all the streets are cleared of snow, the snow plovers are parked on the last intersection they visited. Mirko and Slavko don’t have to finish their plowing on the same intersection.

Write a program that calculates the total amount of fuel that the snow plovers will spend.

Input

The first line of the input contains two integers: N and S, 1 <= N <= 100000, 1 <= S <= N. N is the total number of intersections; S is ordinal number of the snow plovers starting intersection. Intersections are marked with numbers 1...N.

Each of the next N-1 lines contains three integers: A, B and C, meaning that intersections A and B are directly connected by a street and that street's length is C meters, 1 <= C <= 1000.

Output

Write to the output the minimal amount of fuel needed to clean all streets.

Sample Input

5 21 2 12 3 23 4 24 5 1

Sample Output

6

Source

Croatia OI 2002 national – second day, seniors

题目链接：http://poj.org/problem?id=1849

题目大意：有两个人从同一点出发，遍历一棵树的所有边，每个边有个权值，求遍历的最小代价，

题目分析：答案就是所有边权值乘2减去直径，这里不证了，证明的话，简单想就是两个人要走的公共路最长，除了公共路其他部分肯定都走了两次，因为是树，两点间只有唯一的路，所以最长的那条路我不走重复必然就是最小代价了，树上最长链就是树的直径
关于怎么求直径，这里也不证了记个结论，两种方法，一是两次dfs，第一次任意找一个点搜索从它开始能走到的最远的点，那个点必然是树的直径的一个端点，再从这个点开始dfs一下，就搜出了整棵树，二是树形dp的方法，对于任意一个点，搜索其两个能延伸最远和次远的儿子，把两个距离相加再加上两儿子到父亲的距离再取最大就是树的直径，这个很好证，两儿子肯定不在一条延伸路径上，画个图看的更清楚

两次DFS:

#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);}

树形dp：

#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);}