//hdu1879 还是畅通工程
#include <iostream>
#include <algorithm>
#define MAX 5000
using namespace std;
int father[MAX];
int my_rank[MAX];
void make_set()
{
for (int i = 0; i < MAX; i++)
{
my_rank[i] = 0;
father[i] = i;
}
}
int find_set(int x)
{
if (x != father[x])
father[x] = find_set(father[x]);
return father[x];
}
void union_set(int x, int y)
{
x = find_set(x);
y = find_set(y);
if (x == y) return;
if (my_rank[x] > my_rank[y])
father[y] = x;
else
{
father[x] = y;
if (my_rank[x] == my_rank[y])
my_rank[y]++;
}
}
struct edge
{
int begin;
int end;
int weight;
};
int N,M;
edge e[MAX];
bool cmp(const edge& a, const edge& b)
{
return a.weight < b.weight;
}
int main()
{
int count, result;
int x, y;
bool isbuild;
while (cin >> N && N)
{
M = N*(N-1)/2;
make_set();
for (int i = 0; i < M; i++)
{
cin >> e[i].begin >> e[i].end >> e[i].weight >> isbuild;
if (isbuild)
{
x = find_set(e[i].begin);
y = find_set(e[i].end);
if (x != y)
union_set(x, y);
}
}
sort(e, e+M, cmp);
count = 0;
result = 0;
while (count != M)
{
x = find_set(e[count].begin);
y = find_set(e[count].end);
if (x != y)
{
union_set(x, y);
result += e[count].weight;
}
count++;
}
cout << result << endl;
}
return 0;
}
//1863畅通工程
#include<iostream>
#include <algorithm>
#define MAX 105
using namespace std;
int father[MAX];
int my_rank[MAX];
void make_set()
{
for (int i = 0; i < MAX; i++)
{
my_rank[i] = 0;
father[i] = i;
}
}
int find_set(int x)
{
if (x != father[x])
father[x] = find_set(father[x]);
return father[x];
}
void union_set(int x, int y)
{
x = find_set(x);
y = find_set(y);
if (x == y) return;
if (my_rank[x] > my_rank[y])
father[y] = x;
else
{
father[x] = y;
if (my_rank[x] == my_rank[y])
my_rank[y]++;
}
}
struct edge
{
int begin;
int end;
int weight;
};
int N,M;
edge e[MAX];
bool cmp(const edge& a, const edge& b)
{
return a.weight < b.weight;
}
int main()
{
int count, result;
int x, y;
while (cin >> N && N)
{
cin >> M;
for (int i = 0; i < N; i++)
cin >> e[i].begin >> e[i].end >> e[i].weight;
sort(e, e+N, cmp);
count = 0;
result = 0;
make_set();
while (count != N)
{
x = find_set(e[count].begin);
y = find_set(e[count].end);
if (x != y)
{
union_set(x, y);
result += e[count].weight;
}
count++;
}
count = 0;
for (int i = 1; i <= M; i++)
{
if (father[i] == i) count++;
}
if (count == 1)
cout << result << endl;
else
cout << "?" << endl;
}
return 0;
}
//1875畅通工程再续
//这题注意精度问题，要先计算开根号的值，然后在最小生成树时在加起来
//不可先保存没开平方的值，一开始我保存的是没开根号，得到最后答案后再开根号，认为这会准确些
//后来WA卡住了，后来实际想不同就和别人解题报告一比，发现了这个问题
#include <iostream>
#include <algorithm>
#include <math.h>
#include <stdio.h>
#define MAX 6000
using namespace std;
int father[MAX];
int my_rank[MAX];
void make_set()
{
for (int i = 0; i < MAX; i++)
{
father[i] = i;
my_rank[i] = 0;
}
}
int find_set(int x)
{
if (x != father[x])
father[x] = find_set(father[x]);
return father[x];
}
void union_set(int x, int y)
{
x = find_set(x);
y = find_set(y);
if (x == y) return;
if (my_rank[x] > my_rank[y])
father[y] = x;
else
{
father[x] = y;
if (my_rank[x] == my_rank[y])
my_rank[y]++;
}
}
struct coordinate
{
int x, y;
int id;
};
struct edge
{
int begin_id;
int end_id;
double weight;
};
edge e[MAX];
coordinate point[MAX];
bool cmp(const edge& a, const edge& b)
{
return a.weight < b.weight;
}
int main()
{
int cases;
int n;
int count;
double distances;
cin >> cases;
while (cases--)
{
cin >> n;
for (int i = 0; i < n; point[i].id = i,i++)
cin >> point[i].x >> point[i].y;
count = 0;
for (int i = 0; i < n; i++)
{
for (int j = i+1; j < n; j++)
{
distances = sqrt((double)((point[i].x-point[j].x)*(point[i].x-point[j].x) + (point[i].y-point[j].y)*(point[i].y-point[j].y)));
if (distances >= 10.0 && distances <= 1000.0)
{
e[count].begin_id = point[i].id;
e[count].end_id = point[j].id;
e[count].weight = distances;
count++;
}
}
}
sort(e, e+count, cmp);
int m = 0;
make_set();
int x, y;
double result = 0;
while (m != count)
{
x = find_set(e[m].begin_id);
y = find_set(e[m].end_id);
if (x != y)
{
union_set(x, y);
result += e[m].weight;
}
m++;
}
int count = 0;
for (int i = 0; i < n; i++)
if (point[i].id == father[point[i].id])
count++;
if (count == 1)
printf("%.1lf\n", result*100);
else
cout << "oh!" << endl;
}
return 0;
}
