就是问你能否通过选取一些边构成一棵树,最小生成树。
这道题的关键不在于此,在于学到了另外一种优先队列的写法:
struct cmp
{
bool operator ()(Eddge e1, Eddge e2)
{
return e1.val>e2.val;
}
};
priority_queue <Eddge, vector<Eddge>, cmp > Q;
让我的优先队列变得更好看了,并且可以节约大致20~30ms的时间(可以忽略)。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int maxN=505;
int N, M, K, a[maxN], root[maxN];
struct Eddge
{
int nex, to, val;
Eddge(int a=0, int b=0, int c=0):nex(a), to(b), val(c) {}
friend bool operator < (Eddge e1, Eddge e2)
{
return e1.val<e2.val;
}
};
struct cmp
{
bool operator ()(Eddge e1, Eddge e2)
{
return e1.val>e2.val;
}
};
priority_queue <Eddge, vector<Eddge>, cmp > Q;
set<int> st;
void init()
{
st.clear();
for(int i=1; i<=N; i++) root[i]=i;
while(!Q.empty()) Q.pop();
}
int fid(int x)
{
if(root[x] == x) return x;
return root[x] = fid(root[x]);
}
int Kruskal()
{
int ans=0;
while(!Q.empty())
{
Eddge now=Q.top(); Q.pop();
int u=now.nex, v=now.to, cost=now.val;
int rou=fid(u), rov=fid(v);
if(rou!=rov)
{
root[rou]=rov;
ans+=cost;
}
}
for(int i=1; i<=N; i++)
{
st.insert(fid(i));
}
if(st.size()>1) return -1;
return ans;
}
int main()
{
int T; scanf("%d", &T);
while(T--)
{
scanf("%d%d%d", &N, &M, &K);
init();
while(M--)
{
int e1, e2, e3;
scanf("%d%d%d", &e1, &e2, &e3);
Q.push(Eddge(e1, e2, e3));
}
while(K--)
{
int t;
scanf("%d", &t);
for(int i=1; i<=t; i++)
{
scanf("%d", &a[i]);
if(i>1)
{
Q.push(Eddge(a[i-1], a[i], 0));
}
}
}
printf("%d\n", Kruskal());
}
return 0;
}