HDU 4424 and check the collection

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Test instructions: Given to n towns, and then to the value of every two towns, but the value of u->v is the minimum value u->v this path, asking for the maximum value from any point to the other position

Idea: Because the constraint is the minimum value on the path, then we sort by value, from the big to the small row, you can avoid this problem, and then use and check set to complete, for two sets, the value of the left set is a, the value of the right set of the sum of B, now there is a path to join these two sets, So what we're thinking about is merging A into B or merging B into a, so we can compare that the path to the two sets is definitely the smallest value at the moment, and if the number of a sets of elements is Ai, then merging a into B will need to add A*ai, in turn a adds B*bi, and the bigger one. Because they must be merged into a piece, the value of a large left

#pragma COMMENT (linker, "/stack:1024000000,1024000000") #include <stdio.h> #include <string.h> #include <stdlib.h> #include <iostream> #include <algorithm>using namespace std;typedef long Long ll;typedef unsigned long long ull;const int inf=0x3f3f3f3f;const ll inf=0x3f3f3f3f3f3f3f3fll;const int Maxn=200010;int f[maxn];ll nu    m[maxn],r[maxn];struct edge{int u,v; ll cost;} Es[maxn];bool cmp (const edge &a,const edge &b) {return a.cost>b.cost;}    int find1 (int x) {if (x!=f[x]) f[x]=find1 (f[x]); return f[x];}    ll Slove (int n,int k) {for (int i=0;i<=n;i++) num[i]=0;        for (int i=0;i<k;i++) {int u=find1 (ES[I].U);        int V=find1 (ES[I].V);            if (num[u]+r[v]* (es[i].cost) >=num[v]+r[u]* (es[i].cost)) {f[v]=u;num[u]+=r[v]*es[i].cost;        R[U]+=R[V];            }else{F[u]=v;num[v]+=r[u]*es[i].cost;        R[v]+=r[u]; }} return Num[find1 (1)];}    int main () {int n,u,v;    ll cost; while (sCANF ("%d", &n)!=-1) {for (int i=0;i<=n;i++) {f[i]=i;        R[i]=1;        } int k=0;            for (int i=0;i<n-1;i++) {scanf ("%d%d%i64d", &u,&v,&cost);        Es[k].u=u;es[k].v=v;es[k++].cost=cost;        } sort (es,es+k,cmp);        ll Ans=slove (n,k);    printf ("%i64d\n", ans); } return 0;}

HDU 4424 and check the collection