Escape Time II

Time Limit: 20 Sec  Memory Limit: 256 MB

题目连接

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2966

Description

In last winter, there was a big snow storm in South China. The electric system was damaged seriously. Lots of power lines were broken and lots of villages lost contact with the main power grid. The government wants to reconstruct the electric system as soon as possible. So, as a professional programmer, you are asked to write a program to calculate the minimum cost to reconstruct the power lines to make sure there‘s at least one way between every two villages.

Input

Standard input will contain multiple test cases. The first line of the input is a single integer T (1 <= T <= 50) which is the number of test cases. And it will be followed by T consecutive test cases.

In each test case, the first line contains two positive integers N and E (2 <= N <= 500, N <= E <= N * (N - 1) / 2), representing the number of the villages and the number of the original power lines between villages. There follow E lines, and each of them contains three integers, A, B, K (0 <= A, B < N, 0 <= K < 1000). A and B respectively means the index of the starting village and ending village of the power line. If K is 0, it means this line still works fine after the snow storm. If K is a positive integer, it means this line will cost K to reconstruct. There will be at most one line between any two villages, and there will not be any line from one village to itself.

Output

For each test case in the input, there‘s only one line that contains the minimum cost to recover the electric system to make sure that there‘s at least one way between every two villages.

Sample Input

1
3 3
0 1 5
0 2 0
1 2 9

Sample Output

5

HINT

题意

给你个无向边带权图，让你输出最小生成树

题解：

Kruskal or prim

代码：

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
#define maxn 200001
#define mod 10007
#define eps 1e-9
int Num;
char CH[20];
//const int inf=0x7fffffff;   //нчоч╢С
const int inf=0x3f3f3f3f;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
inline void P(int x)
{
    Num=0;if(!x){putchar(‘0‘);puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
//**************************************************************************************

int u[maxn],v[maxn],w[maxn],r[maxn];
int pa[maxn],n,m;
bool cmp(int i,int j)
{
    return w[i]<w[j];
}
int fi(int x)
{
    return pa[x]==x?x:pa[x]=fi(pa[x]);
}
int kruskal()
{
    int ans=0;
    for(int i=0;i<n;i++)
        pa[i]=i;
    for(int i=0;i<m;i++)
        r[i]=i;
    sort(r,r+m,cmp);
    for(int i=0;i<m;i++)
    {
        int e=r[i];
        int x=fi(u[e]);
        int y=fi(v[e]);
        if(x!=y)
            ans+=w[e],pa[x]=y;
    }
    return ans;
}
int main()
{
    //test;
    int t=read();
    while(t--)
    {
        n=read(),m=read();
        for(int i=0;i<m;i++)
        {
            int a=read(),b=read(),c=read();
            u[i]=a,v[i]=b,w[i]=c;
        }
        printf("%d\n",kruskal());
    }
}