HDU 4679 Terrorist’s destroy (2013多校8 1004题 树形DP)

Terrorist’s destroy

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 24    Accepted Submission(s): 6

Problem Description

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.

 

Input

The 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)

 

Output

For 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

 

Source

2013 Multi-University Training Contest 8

 

Recommend

zhuyuanchen520

各种dfs,导致爆栈了，加个栈挂，C++交果断AC了。

我是先求树的直径。

如果去掉的不是树的直径上的边，那么去掉后乘积就是w*直径长度

如果去掉直径上的边，那么直径两端各dfs一次就可以了

/* ***********************************************
Author        :kuangbin
Created Time  :2013/8/15 14:48:54
File Name     :F:2013ACM练习2013多校81004.cpp
************************************************ */
#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;

const int MAXN = 100010;
struct Edge
{
    int to,next;
    int id;
    int w;
}edge[MAXN*2];
int mm[MAXN];
int maxn[MAXN];
int smaxn[MAXN];
int head[MAXN],tot;
void init()
{
    memset(head,-1,sizeof(head));
    tot = 0;
}
void addedge(int u,int v,int w,int id)
{
    edge[tot].to = v;
    edge[tot].w = w;
    edge[tot].id = id;
    edge[tot].next = head[u];
    head[u] = tot++;
    edge[tot].to = u;
    edge[tot].w = w;
    edge[tot].id = id;
    edge[tot].next = head[v];
    head[v] = tot++;
}
void dfs(int u,int pre)
{
    mm[u] = 0;
    maxn[u] = 0;
    smaxn[u] = 0;
    for(int i = head[u];i != -1;i = edge[i].next)
    {
        int v = edge[i].to;
        if(v == pre)continue;
        dfs(v,u);
        if(maxn[v]+1 > smaxn[u])
        {
            smaxn[u] = maxn[v] + 1;
            if(smaxn[u] > maxn[u])
            {
                swap(smaxn[u],maxn[u]);
            }
        }
        if(mm[v] > mm[u])
            mm[u] = mm[v];
    }
    mm[u] = max(mm[u],maxn[u]+smaxn[u]);
}
int ans;
int dep[MAXN];
int p[MAXN];
bool used[MAXN];
int cnt;
int index;
int a[MAXN];
void solve(int u,int pre)
{
    for(int i = head[u];i != -1;i = edge[i].next)
    {
        int v = edge[i].to;
        int w = edge[i].w;
        if(v == pre)continue;
        solve(v,u);
        if(used[v])
        {
            a[edge[i].id] = max(a[edge[i].id],w*mm[v]);
        }
        else
        {
            a[edge[i].id] = max(a[edge[i].id],w*cnt);
        }
    }
}
;
void dfs1(int u,int pre)
{
    p[u] = pre;
    dep[u] = dep[pre] + 1;
    for(int i = head[u]; i != -1;i = edge[i].next)
    {
        int v = edge[i].to;
        if(v==pre)continue;
        dfs1(v,u);
    }
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int T;
    int n;
    scanf("%d",&T);
    int u,v,w;
    int iCase = 0;
    while(T--)
    {
        iCase ++;
        init();
        scanf("%d",&n);
        for(int i = 1;i < n;i++)
        {
            scanf("%d%d%d",&u,&v,&w);
            addedge(u,v,w,i);
        }
        dep[0] = 0;
        dfs1(1,0);
        u = 1;
        for(int i = 1;i <= n;i++)
            if(dep[u] < dep[i])
                u = i;
        dfs1(u,0);
        v = 1;
        for(int i =1;i <= n;i++)
            if(dep[v] < dep[i])
                v = i;
        cnt = dep[v]-1;
        memset(used,false,sizeof(used));
        int tmp = v;
        while(tmp)
        {
            used[tmp] = true;
            tmp = p[tmp];
        }
        for(int i = 1;i <= n;i++)
            a[i] = 0;
        ans = 1000000000;
        dfs(u,0);
        solve(u,-1);
        dfs(v,0);
        solve(v,-1);
        for(int i = 1;i < n;i++)
            if(a[i]<ans)
            {
                ans = a[i];
                index = i;
            }
        printf("Case #%d: %d
",iCase,index);
    }
    
    return 0;
}