Nowadays the one-way traffic is introduced all over the world in order to improve driving safety and reduce traffic jams. The government of Dhaka Division decided to keep up with new trends. Formerly all n cities of Dhaka were connected by n two-way roads in the ring, i.e. each city was connected directly to exactly two other cities, and from each city it was possible to get to any other city. Government of Dhaka introduced one-way traffic on all n roads, but it soon became clear that it's impossible to get from some of the cities to some others. Now for each road is known in which direction the traffic is directed at it, and the cost of redirecting the traffic. What is the smallest amount of money the government should spend on the redirecting of roads so that from every city you can get to any other?
Input
Input starts with an integer T (≤ 200), denoting the number of test cases.
Each case starts with a blank line and an integer n (3 ≤ n ≤ 100) denoting the number of cities (and roads). Next n lines contain description of roads. Each road is described by three integers ai, bi, ci (1≤ai,bi≤n,ai≠bi,1≤ci≤100) - road is directed from city ai to city bi, redirecting the traffic costs ci.
Output
For each case of input you have to print the case number and the smallest amount of money the government should spend on the redirecting of roads so that from every city you can get to any other.
Sample Input
4
3
1 3 1
1 2 1
3 2 1
3
1 3 1
1 2 5
3 2 1
6
1 5 4
5 3 8
2 4 15
1 6 16
2 3 23
4 6 42
4
1 2 9
2 3 8
3 4 7
4 1 5
Output for Sample Input
Case 1: 1
Case 2: 2
Case 3: 39
Case 4: 0
题目大意就是说n个点有n条单项边连接（如果将单向边看成一条线即是一个环），如果让任意一个点都可以到达其他点，求最小花费。调整一个单项边的方向花费的是这条边的权值。
那么只有两种情况顺时针和逆时针，将两者中最小的花费输出就好了。
#include<bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
bool visit[107];
int dfs(vector<vector<pair<int,int> > >&vec, int n)
{
visit[n] = true;
int res = 0;
if(visit[vec[n][0].first] == false)
res = vec[n][0].second + dfs(vec, vec[n][0].first);
if(visit[vec[n][1].first] == false)
res = vec[n][1].second + dfs(vec, vec[n][1].first);
return res;
}
void solve(int cases)
{
int n;
scanf("%d", &n);
vector<vector<pair<int,int> > >vec(n+1);
int a, b, c;
for(int i=0; i<n; ++ i)
{
scanf("%d%d%d", &a, &b, &c);
vec[a].push_back(make_pair(b, 0));
vec[b].push_back(make_pair(a, c));
}
memset(visit, false, sizeof(visit));
visit[1] = true;
int res1 = vec[1][0].second + dfs(vec, vec[1][0].first);
int t1 = vec[1][1].first;
if(vec[t1][0].first == 1)
res1 += vec[t1][0].second;
else
res1 += vec[t1][1].second;
memset(visit, false, sizeof(visit));
visit[1] = true;
int res2 = vec[1][1].second + dfs(vec, vec[1][1].first);
int t2 = vec[1][0].first;
if(vec[t2][0].first == 1)
res2 += vec[t2][0].second;
else
res2 += vec[t2][1].second;
printf("Case %d: %d
", cases, min(res1, res2));
}
int main()
{
int t;
scanf("%d", &t);
for(int i=1; i<=t; ++ i)
solve(i);
return 0;
}