UVA, 11733 （kruscal，最小生成树）

Airports

From:UVA,
11733

Submit

Time Limit: 1000 MS

The government of a certain developing nation wants to improve transportation
in one of its most inaccessible areas, in an attempt to attract investment. The region consists of several important locations that must have access to an airport.

Of course, one option is to build an airport in each of these places, but it may turn out to be cheaper to build fewer airports and have roads link them to all of the other locations. Since these are
long distance roads connecting major locations in the country (e.g. cities, large villages, industrial areas), all roads are two-way. Also, there may be more than one direct road possible between two areas. This is because there may be several ways to link
two areas (e.g. one road tunnels through a mountain while the other goes around it etc.) with possibly differing costs.

A location is considered to have access to an airport either if it contains an airport or if it is possible to travel by road to another location from there that has an airport.

You are given the cost of building an airport and a list of possible roads between pairs of locations and their corresponding costs. The government now needs your help to decide on the cheapest way
of ensuring that every location has access to an airport. The aim is to make airport access as easy as possible, so if there are several ways of getting the minimal cost, choose the one that has the most airports.

Note: The input file is large; make sure your I/O code is fast.

Input

The first line of input contains the integer T(T<25), the number of test cases. The rest of the input consists of T cases.

Each case starts with three integers N, M and A (0<N<=10,000, 0<=M<=100,000, 0<A<=10,000) separated
by white space. N is the number of locations, M is the number of possible roads that can be built, and A is the cost of building an airport.

The following M lines each contain three integers X, Y and C (1<=X, Y<=N, 0<C<=10,000),
separated by white space. X and Y are two locations, and C is the cost of building a road between X and Y.

Output

Your program should output exactly T lines, one for each case. Each line should be of the form "Case #X: Y Z", where X is the case number Y is
the minimum cost of making roads and airports so that all locations have access to at least one airport, and Z is the number of airports to be built. As mentioned earlier, if there are several answers with minimal cost, choose the one that
maximizes the number of airports.

题目大意：

有N个城镇，m条马路，需要建造飞机场，建造马路和飞机场的费用告诉你，让你输出在村庄都可达飞机场的情况下最小花费和在最小花费下所能建造的最多的飞机场的数目。

解题思路：

kruscal算法，对边排序，当边的费用大于飞机场的费用时，舍弃边，建造飞机场数目cnt++，最后飞机场的cnt=cnt+集合的个数。算出费用和。

代码：

#include<iostream>
#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;

struct edge{int u,v,cost;};

const int maxm=100010,maxn=10010;

bool cmp(const edge& e1,const edge& e2){
    return e1.cost<e2.cost;
}

edge ed[maxm];
int v,e,air,parent[maxn],cnt,res,rank[maxn];
vector <int> myve;

void input(){
    scanf("%d%d%d",&v,&e,&air);
    for(int i=0;i<e;i++){
        scanf("%d %d %d",&ed[i].u,&ed[i].v,&ed[i].cost);
    }
}

void inti_union_find(int v){
    for(int i=0;i<=v;i++){
        parent[i]=i;
        rank[i]=1;
    }
}

void Union(int a,int b){
    if(rank[a]<rank[b]){
        parent[a]=b;
    }
    else{
        if(rank[a]==rank[b]) rank[a]++;
        parent[b]=a;
    }
}

int find(int x){return parent[x]==x?x:parent[x]=find(parent[x]);}

void kruskal(){
    sort(ed,ed+e,cmp);
    inti_union_find(v);
    res=0;
    cnt=0;
    for(int i=0;i<e;i++){
        edge tempe=ed[i];
        int p=find(tempe.u);
        int q=find(tempe.v);
        if(q!=p){
            Union(p,q);
            res+=tempe.cost;
            if(tempe.cost>=air){
                res-=tempe.cost;
                cnt++;
            }
        }
    }
}

void solve(){
    for(int i=1;i<=v;i++){
        if(find(i)==i)
            cnt++;
    }
    res+=air*cnt;
}

int main(){
    int t,casen=0;
    scanf("%d",&t);
    while(t--){
        input();
        kruskal();
        solve();
        printf("Case #%d: %d %d\n",++casen,res,cnt);
    }
    return 0;
}