数据结构 通畅工程 (最小生成树)

Description

在n个城市之间建设网络，只需保证连通即可，求最经济的架设方法。

Input

有多组输入数据。每组第一行输入三个整数n、m、c（1<=n,m,c<=100000），分别代表城市数量，可建道路数量和单位长度道路修建费用。接下来m行每行三个整数u、v(1<=u,v<=n)、d(1<=d<=100000)。代表可建道路的起点城市、终点城市和长度。

Output

每组数据输出一行，输出数据组数和使所有城市连通的最小费用，无法全部连通输出-1，格式见样例。

Sample Input

1 2 1 100 1 2 100

Sample Output

Case #1: 10000

HINT

考察知识点：最小生成树。

Append Code

析：没什么可说的，最小生成树，水题。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define debug puts("+++++")
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;
 
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 5;
const LL mod = 1e9 + 7;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
inline int gcd(int a, int b){  return b == 0 ? a : gcd(b, a%b); }
inline int lcm(int a, int b){  return a * b / gcd(a, b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
struct Node{
    int u, v;
    LL val;
    bool operator < (const Node &p) const{
        return val < p.val;
    }
};
Node a[maxn];
int p[maxn];
int Find(int x){  return x == p[x] ? x : p[x] = Find(p[x]); }
LL k;
 
LL Kruskal(){
    int cnt = 0;
    LL ans = 0;
    for(int i = 0; i < m && cnt < n-1; ++i){
        int x = Find(a[i].u);
        int y = Find(a[i].v);
        if(x != y){
            p[y] = x;
            ans += a[i].val;
            ++cnt;
        }
    }
    ans *= k;
    return cnt == n-1 ? ans : -1;
}
 
int main(){
    int T;  cin >> T;
    for(int kase = 1; kase <= T; ++kase){
        scanf("%d %d %lld", &n, &m, &k);
        for(int i = 0; i <= n; ++i)  p[i] = i;
        for(int i = 0; i < m; ++i)
            scanf("%d %d %lld", &a[i].u, &a[i].v, &a[i].val);
        sort(a, a+m);
        printf("Case #%d: %lld
", kase, Kruskal());
    }
    return 0;
}