树形DP 2013多校8（Terrorist’s destroy HDU4679）

题意：

There is a city which is built like a tree.A terrorist wants to destroy the city's roads. But now he is alone, he can only destroy one road, then the city will be divided into two cities. Impression of the city is a number defined as the distance between the farthest two houses (As it relates to the fare).When the terrorist destroyed a road, he needs to spend some energy, assuming that the number is a.At the same time,he will get a number b which is maximum of the Impression of two cities. The terrorist wants to know which road to destroy so that the product of a and b will be minimized.You should find the road's id.
Note that the length of each road is one.

给出一个树形图，和边权值w代表耗费的能量，每条边的长度是1；破坏一条边<u,v>会形成两棵树，b是两个树中较大的那个直径，a是<u,v>的能量值，求破坏哪条边保证a*b的至最小若存在多个，则输出最先出现的那条边；

分析：求树的直径，我们很容易想到两个dfs可以求出树的直径，所以先树形dp（前两个dfs）求出每个点的正向距离的最大值dis[u][0]和次大值dis[u][1]，以及反向距离最大值dis[u][2]，然后第三个dfs深搜枚举每条边的两个点<u,v>,对于v点，v所对应的子树的直径就是

dis[v][0]+dis[v][1];

对于v如果belong[u]==v这u所对应的子树的直径是dis[u][1]+dis[u][2];否则对应的子树直径是dis[u][0]+dis[u][2];

然后枚举a*b即可：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include"stdio.h"
#include"string.h"
#include"stdlib.h"
#include"queue"
#include"algorithm"
#include"string.h"
#include"string"
#include"math.h"
#include"vector"
#include"stack"
#include"map"
#define eps 1e-4
#define inf 0x3f3f3f3f
#define M 100009
#define PI acos(-1.0)
using namespace std;
struct node
{
    int u,v,w,next;
}edge[M*2];
int t,head[M],dis[M][4],length[M*2],belong[M];
__int64 ans,num[M],n;
void init()
{
    t=0;
    memset(head,-1,sizeof(head));
}
void add(int u,int v,int w)
{
    edge[t].u=u;
    edge[t].v=v;
    edge[t].w=w;
    edge[t].next=head[u];
    head[u]=t++;
}
void dfs(int u,int f)
{
    dis[u][0]=dis[u][1]=dis[u][2]=0;
    for(int i=head[u];~i;i=edge[i].next)
    {
        int v=edge[i].v;
        if(v==f)continue;
        dfs(v,u);
        if(dis[u][0]<dis[v][0]+1)
        {
            dis[u][1]=dis[u][0];
            dis[u][0]=dis[v][0]+1;
            belong[u]=v;
        }
        else if(dis[u][1]<dis[v][0]+1)
            dis[u][1]=dis[v][0]+1;
    }
}
void dfs1(int u,int f)
{
    for(int i=head[u];i!=-1;i=edge[i].next)
    {
        int v=edge[i].v;
        if(v==f)
            continue;
        if(belong[u]==v)
            dis[v][2]=max(dis[u][1],dis[u][2])+1;
        else
            dis[v][2]=max(dis[u][0],dis[u][2])+1;
        dfs1(v,u);
    }
}
void dfs2(int u,int f)
{
    for(int i=head[u];~i;i=edge[i].next)
    {
        int v=edge[i].v;
        if(v==f)continue;
        if(belong[u]==v)
        {
            length[i]=dis[v][0]+dis[v][1];
            length[i^1]=dis[u][1]+dis[u][2];
        }
        else
        {
            length[i]=dis[v][0]+dis[v][1];
            length[i^1]=dis[u][0]+dis[u][2];
        }
        dfs2(v,u);
    }
}
int main()
{
    int Case,i,n,a,b,c,kk=1;
    scanf("%d",&Case);
    while(Case--)
    {
        scanf("%d",&n);
        init();
        for(i=1;i<n;i++)
        {
            scanf("%d%d%d",&a,&b,&c);
            add(a,b,c);
            add(b,a,c);
        }
        dfs(1,-1);
        dfs1(1,-1);
        dfs2(1,-1);
        ans=inf;
        int id;
        for(i=0;i<t;i+=2)
        {
            //printf("%d %d %d %d
",edge[i].u,edge[i].v,length[i],length[i^1]);
            int m=max(length[i],length[i^1])*edge[i].w;
            if(ans>m)
            {
                id=i;
                ans=m;
            }
        }
        printf("Case #%d: %d
",kk++,(id+2)/2);
    }
}