Conquer a New Region
Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 657 Accepted Submission(s): 179
Problem Description
The wheel of the history rolling forward, our king conquered a new region in a distant continent.
There are N towns (numbered from 1 to N) in this region connected by several roads. It's confirmed that there is exact one route between any two towns. Traffic is important while controlled colonies are far away from the local country. We define the capacity C(i, j) of a road indicating it is allowed to transport at most C(i, j) goods between town i and town j if there is a road between them. And for a route between i and j, we define a value S(i, j) indicating the maximum traffic capacity between i and j which is equal to the minimum capacity of the roads on the route.
Our king wants to select a center town to restore his war-resources in which the total traffic capacities from the center to the other N - 1 towns is maximized. Now, you, the best programmer in the kingdom, should help our king to select this center.
There are N towns (numbered from 1 to N) in this region connected by several roads. It's confirmed that there is exact one route between any two towns. Traffic is important while controlled colonies are far away from the local country. We define the capacity C(i, j) of a road indicating it is allowed to transport at most C(i, j) goods between town i and town j if there is a road between them. And for a route between i and j, we define a value S(i, j) indicating the maximum traffic capacity between i and j which is equal to the minimum capacity of the roads on the route.
Our king wants to select a center town to restore his war-resources in which the total traffic capacities from the center to the other N - 1 towns is maximized. Now, you, the best programmer in the kingdom, should help our king to select this center.
Input
There are multiple test cases.
The first line of each case contains an integer N. (1 <= N <= 200,000)
The next N - 1 lines each contains three integers a, b, c indicating there is a road between town a and town b whose capacity is c. (1 <= a, b <= N, 1 <= c <= 100,000)
The first line of each case contains an integer N. (1 <= N <= 200,000)
The next N - 1 lines each contains three integers a, b, c indicating there is a road between town a and town b whose capacity is c. (1 <= a, b <= N, 1 <= c <= 100,000)
Output
For each test case, output an integer indicating the total traffic capacity of the chosen center town.
Sample Input
4
1 2 2
2 4 1
2 3 1
4
1 2 1
2 4 1
2 3 1
Sample Output
4
3
Source
Recommend
zhuyuanchen520
题意：
给出一棵树，找出一个点，求出所有点到这个点的权值和最大，
权值为路径上所有边权的最小值。
分析：
由于要所有点到这个点的权值和最大，把边按从大到小排序并插入。每条边连接两个集合，
且每次并入的边权值都是当前已并入边中最小的。那么，只要每次并入时判断是把a并入b
得到的权值和大还是b并入a得到的权值和大就可以了。并查集维护集合的元素个数和总的
权值。
事实上，边总会并入最大边所在的集合，为什么呢？
可以用不等式说明,比如：排序后 三条边的权值分别为x1>x2>x3
那么并x3时，如果是把x1和x2已并的集合加到x3这个集合中来，必须满足：
x1+x2(即sum)+x3<3*x3，化简得X3>(X1+X2)/2，而x2<x1,那么x3>x2（与假设矛盾，不成立）
因此，每次边的并入都是加再最大边所在的集合中。
感想：
1、首先读题的时候，对最大权值 和 权值取路径上所有边权的最小值 理解混淆，"最大"、"最小"。。。
导致样例都没看明白啊啊啊。。。
正确理解：
那个距离其实就是容量，可以理解为路最大的承载，如果大于那个值对应的路段会倒掉。。
那么你开车从1到3，车上最多的容量为1。。
2、完了以后再hdu oj上A的代码在zoj 上wa，最后发现是输出语句中，要把"%I64d"换成“%lld".
3、A完后，我又想既然直接就是并入最大边所在的集合，那不就直接把边都加起来就行了，事实证明不行啊。
为什么呢?因为。。。不一定都像input给出的数据都是树啊，如果是图就有问题了。4个点，3条边，如果
一个点孤立。因此必须要用并查集维护啊= =不对。。。题意只能是树，那又为什么不能直接加呢?求解！！
代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#define N 200010
typedef long long ll;
using namespace std;
struct node
{
int u,v,w;
bool operator <(const node a)const
{
return w>a.w;
}
}edge[N];
int cnt[N],pre[N]; //pre[]记录前一个节点编号
ll sum[N]; //sum[i]表示i为根的边权和，cnt[i]记录i为根的树中元素个数
int find(int a) //找根节点
{
return pre[a]=(pre[a]==a?a:find(pre[a]));
}
int main()
{
int n,i;
while(scanf("%d",&n)!=EOF)
{
for(i=1;i<=n-1;i++)
{
scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].w);
}
sort(edge+1,edge+n); //从大到小排序
for(i=1;i<=n;i++)
{
cnt[i]=1;
sum[i]=0;
pre[i]=i;
}
ll ans=0;
for(i=1;i<=n-1;i++)
{
int ra=find(edge[i].u);
int rb=find(edge[i].v);
ll bisr=sum[rb]+(ll)edge[i].w*cnt[ra];
ll aisr=sum[ra]+(ll)edge[i].w*cnt[rb];
if(bisr>aisr)
{
pre[ra]=rb;
sum[rb]=bisr;
cnt[rb]+=cnt[ra];
}
else
{
pre[rb]=ra;
sum[ra]=aisr;
cnt[ra]+=cnt[rb];
}
ans=max(ans,max(aisr,bisr));
}
//printf("%I64d
",ans);
printf("%lld
",ans); //zoj上需要这么写才能AC
}
return 0;
}
//bisr=把a并入b,aisr=把b并入a
//注意要用long long