Zoj 3659 Conquer a New Region and query set + greedy, zoj3659

Zoj 3659 Conquer a New Region and query set + greedy, zoj3659

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Conquer a New Region Time Limit: 5 Seconds Memory Limit: 32768 KB

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, shocould 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)

Output

For each test case, output an integer indicating the total traffic capacity of the chosen center town.

Sample Input

41 2 22 4 12 3 141 2 12 4 12 3 1

Sample Output

43

Contest: The 2012 ACM-ICPC Asia Changchun Regional Contest

Listen to students saying that they are about the water in the semi-finals

Is this the legendary quiz question?

Well, I don't know how to solve it.

There is a feeling of "Maximum spanning tree"

However, Baidu does not have this question for the maximum spanning tree.

N-1 edges and ask the sum of edge weights sent from one vertex to another (the edge weights sent to other vertices are determined by the smallest edge weights on the path)

Train of Thought: first sort by given side from large to small and then greedy

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;struct Edge{    int s,t;    int len;}edge[200200];int cmp(Edge a,Edge b){    return a.len>b.len;}struct Node{    int all;    long long int sum;}node[200200];int father[200200];int find(int x){    if(father[x]!=x)        father[x]=find(father[x]);    return father[x];}int n;int main(){    int i;    long long int ans;    while(scanf("%d",&n)!=EOF)    {        ans=0;        for(i=1;i<n;i++)        {            scanf("%d %d %d",&edge[i].s,&edge[i].t,&edge[i].len);        }        sort(edge+1,edge+n,cmp);        for(i=1;i<=n;i++)        {            father[i]=i;            node[i].all=1;            node[i].sum=0;        }        for(i=1;i<n;i++)        {            int fa=find(edge[i].s);            int fb=find(edge[i].t);            long long int s1=node[fb].sum+(long long int)edge[i].len*node[fa].all;            long long int s2=node[fa].sum+(long long int)edge[i].len*node[fb].all;            if(s1>=s2)            {                node[fa].sum=s1;                node[fa].all+=node[fb].all;                father[fb]=fa;                if(ans<s1)                    ans=s1;            }            else            {                node[fb].sum=s2;                node[fb].all+=node[fa].all;                father[fa]=fb;                if(ans<s2)                    ans=s2;            }        }        printf("%lld\n",ans);    }    return 0;}