Terrorist’s destroy HDU - 4679

Posted 2021-02-05 smallhester

Terrorist’s destroy HDU - 4679 

 

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. 

InputThe first line contains integer T(1<=T<=20), denote the number of the test cases. 
For each test cases,the first line contains a integer n(1 < n <= 100000);denote the number of the houses; 
Each of the following (n-1) lines contains third integers u,v,w, indicating there is a road between house u and houses v,and will cost terrorist w energy to destroy it.The id of these road is number from 1 to n-1.(1<=u<=n , 1<=v<=n , 1<=w<=10000)OutputFor each test case, output the case number first,and then output the id of the road which the terrorist should destroy.If the answer is not unique,output the smallest id.Sample Input

2
5
4 5 1
1 5 1
2 1 1
3 5 1
5
1 4 1
1 3 1
5 1 1
2 5 1

Sample Output

Case #1: 2
Case #2: 3


题意：给你一棵树和边上的权值，定义去掉一条边的花费为边权值（a）乘上b，b定义为去掉边后形成的两棵树中两点间的最远距离（注意是各自内部的最远距离），问去掉哪条边的花费最少？

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<stack>
#include<cstdlib>
#include <vector>
#include<queue>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 200020;
int q[maxn],dislong[maxn],deleteLeft[maxn],deleteRight[maxn],pre[maxn],father[maxn],dep[maxn];
int list[maxn],Next[maxn],p[maxn],c[maxn],id[maxn],maxx,n;
bool b[maxn],inlong[maxn];


int findlong(int xx,int n)
{
    int t,w,now,k,x;
    for (int i = 1; i <= n; i++)
    {
        b[i] = true;
    }
    t = 0; w = 1;
    q[1] = xx;
    dislong[1] = 1;
    b[xx] = false;
    pre[xx] = 0;
    maxx = 0;
    while (t < w)
    {
        t++; x = q[t];
        k = list[x];
        while (k > 0)
        {
            if (b[p[k]] == true)
            {
                w++;
                b[p[k]] = false;
                q[w] = p[k];
                dislong[w] = dislong[t]+1;
                pre[p[k]] = x;
                if (dislong[w] > maxx) {maxx = dislong[w]; now = p[k];}
            }
            k = Next[k];
        }
    }
    return now;
}
void init(int n)
{
    for (int i = 1; i <= n; i++)
    {
        list[i] = 0;
    }
}
void dfs_dep(int x,int pre1)
{
    int k;
    dep[x] = 1;
    k = list[x];
    while (k > 0)
    {
        if (inlong[p[k]] == false && p[k] != pre1)
        {
            dfs_dep(p[k],x);
            dep[x] = max(dep[x],dep[p[k]]+1);
        }
        k = Next[k];
    }
}

int main()
{
    int t;
    scanf("%d",&t);
   int ca = 1;
   while(t--)
   {
        int tot = 0;
        scanf("%d",&n);
        init(n);
        for (int i = 1;i < n; i++)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            tot++;
            Next[tot] = list[u];
            list[u] = tot;
            p[tot] = v;
            c[tot] = w;
            id[tot] = i;
            tot++;
            Next[tot] = list[v];
            list[v] = tot;
            p[tot] = u;
            c[tot] = w;
            id[tot] = i;
        }
        if (n != 1)
        {
            memset(pre,0,sizeof(pre));
            int front = findlong(1,n);
            int rear = findlong(front,n);
            int sum = maxx-1;

            int k = rear;
            memset(inlong,false,sizeof(inlong));
            memset(father,0,sizeof(father));
            while (k > 0)
            {
                inlong[k] = true;
                father[pre[k]] = k;
                k = pre[k];
            }
            memset(dep,0,sizeof(dep));
            for(int i=1;i<=n;i++)
                if(inlong[i])
                    dfs_dep(i,0);
            memset(deleteLeft,0,sizeof deleteLeft);
            memset(deleteRight,0,sizeof deleteRight);
            k = front;
            int step = 0;
            while (k != rear)
            {
                step++;
                deleteLeft[k] = max(deleteLeft[pre[k]],step-1+dep[k]-1);
                k = father[k];
            }

            k = rear; step = 0;
            father[rear] = 0;
            while (k != front)
            {
                step++;
                deleteRight[k] = max(deleteRight[father[k]],step-1+dep[k]-1);
                k = pre[k];
            }
            //遍历直径
            int ans = INF;
            int result = INF;
            k = front;
            int kk;
            while (k != rear)
            {
                kk = list[k];
                while (kk > 0)
                {
                    if (p[kk] == father[k]) break;
                    kk = Next[kk];
                }
                if (ans > c[kk]*max(deleteLeft[k],deleteRight[father[k]]))
                {
                    ans = c[kk]*max(deleteLeft[k],deleteRight[father[k]]);
                    result = id[kk];

                }
                else if (ans == c[kk]*max(deleteLeft[k],deleteRight[father[k]]))
                {
                    if (result > id[kk]) result = id[kk];
                }
                k = father[k];
            }

            //遍历枝条
            for (int i = 1; i <= n; i++)
            {
                k = list[i];
                while (k > 0)
                {
                    if(!(inlong[i] && inlong[p[k]]))
                    {
                        if (ans > c[k]*sum)
                        {
                            ans = c[k]*sum;
                            result = id[k];

                        }
                        else if (ans == c[k]*sum)
                        {
                            if (result > id[k]) result = id[k];
                        }
                    }
                    k = Next[k];
                }
            }
            printf("Case #%d: %d
",ca++,result);

        }
    }
    return 0;
}

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