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In order to prepare the “The First National ACM School Contest”(in 20??) the major of the city decided to provide all the schools with a reliable source of power. (The major is really afraid of blackoutsJ). So, in order to do that, power station “Future” and one school (doesn’t matter which one) must be connected; in addition, some schools must be connected as well.
You may assume that a school has a reliable source of power if it’s connected directly to “Future”, or to any other school that has a reliable source of power. You are given the cost of connection between some schools. The major has decided to pick out two the cheapest connection plans – the cost of the connection is equal to the sum of the connections between the schools. Your task is to help the major – find the cost of the two cheapest connection plans.
Input
The Input starts with the number of test cases, T (1£T£15) on a line. Then T test cases follow. The first line of every test case contains two numbers, which are separated by a space, N (3£N£100) the number of schools in the city, and M the number of possible connections among them. Next M lines contain three numbers Ai, Bi, Ci , where Ci is the cost of the connection (1£Ci£300) between schools Ai and Bi. The schools are numbered with integers in the range 1 to N.
Output
For every test case print only one line of output. This line should contain two numbers separated by a single space - the cost of two the cheapest connection plans. Let S1 be the cheapest cost and S2 the next cheapest cost. It’s important, that S1=S2 if and only if there are two cheapest plans, otherwise S1£S2. You can assume that it is always possible to find the costs S1 and S2..
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Sample Input |
Sample Output |
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2 5 8 1 3 75 3 4 51 2 4 19 3 2 95 2 5 42 5 4 31 1 2 9 3 5 66 9 14 1 2 4 1 8 8 2 8 11 3 2 8 8 9 7 8 7 1 7 9 6 9 3 2 3 4 7 3 6 4 7 6 2 4 6 14 4 5 9 5 6 10 |
110 121 37 37 |
Problem source: Ukrainian National Olympiad in Informatics 2001
Problem author: Shamil Yagiyayev
Problem submitter: Dmytro Chernysh
Problem solution: Shamil Yagiyayev, Dmytro Chernysh, K M Hasan
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一个裸的次小生成树，没什么好说的，我肯定做麻烦了
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#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=211;
const int maxm=21111;
const double eps=1e-12;
const double OO=1e100;
struct Road{
int u,v;
int w;
Road(){}
Road(int u,int v,int w){
this->u=u;
this->v=v;
this->w=w;
}
bool operator<(const Road& rhs) const{
return w<rhs.w;
}
};
class DisjointSet{
private:
int pa[maxn];
int n;
public:
void makeSet(int n){
this->n=n;
for (int i=0;i<=n;i++) pa[i]=i;
}
int findSet(int x){
if (x!=pa[x]) pa[x]=findSet(pa[x]);
return pa[x];
}
void unionSet(int x,int y){
x=findSet(x);
y=findSet(y);
if (x!=y) pa[x]=y;
}
}disjointSet;
class Graph{
private:
struct EdgeNode{
int to;
int w;
int next;
};
int head[maxn],edge,n;
EdgeNode edges[maxm*2];
int maxCost[maxn][maxn];
void dfsCost(int s,int u,int pa){
for (int i=head[u];i!=-1;i=edges[i].next){
int v=edges[i].to;
int w=edges[i].w;
if (v==pa) continue;
maxCost[s][v]=maxCost[s][u];
if (maxCost[s][v]<w) maxCost[s][v]=w;
dfsCost(s,v,u);
}
}
public:
void init(int n){
this->n=n;
memset(head,-1,sizeof(int)*(n+1));
edge=0;
}
void addedge(int u,int v,int w){
edges[edge].w=w,edges[edge].to=v,edges[edge].next=head[u],head[u]=edge++;
}
void findMaxCost(){
for (int i=1;i<=n;i++)
for (int j=1;j<=n;j++)
maxCost[i][j]=0;
for (int i=1;i<=n;i++){
dfsCost(i,i,-1);
}
}
int getMaxCost(int x,int y){
return maxCost[x][y];
}
}tree;
vector<Road>road;
int n,m;
int minTree;
bool can[maxn][maxn];
void initializer(){
memset(can,0,sizeof(can));
disjointSet.makeSet(n);
road.clear();
minTree=0;
tree.init(n);
}
void input(){
for (int i=0;i<m;i++){
int a,b,w;
scanf("%d%d%d",&a,&b,&w);
road.push_back(Road(a,b,w));
}
}
void Kruskal(){
sort(road.begin(),road.end());
for (vector<Road>::iterator it=road.begin();it!=road.end();it++){
int u=it->u;
int v=it->v;
int w=it->w;
if (disjointSet.findSet(u)!=disjointSet.findSet(v)){
minTree+=w;
disjointSet.unionSet(u,v);
tree.addedge(u,v,w);
tree.addedge(v,u,w);
can[u][v]=can[v][u]=true;
}
}
tree.findMaxCost();
}
void solve(){
int ans=INF;
for (vector<Road>::iterator it=road.begin();it!=road.end();it++){
int u=it->u;
int v=it->v;
int w=it->w;
if (!can[u][v]){
ans=min(ans,minTree-tree.getMaxCost(u,v)+w);
}
}
printf("%d %d
",minTree,ans);
}
int main()
{
int T;
scanf("%d",&T);
while (T--){
scanf("%d%d",&n,&m);
initializer();
input();
Kruskal();
solve();
}
return 0;
}
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