【DFS+题意】#24 A. Ring road

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本文探讨了在交通网络中引入单向行驶后如何通过最小化成本重新定向道路以确保从任何一个城市到另一个城市的通行路径。通过算法实现，提供了解决方案以优化城市间的交通流动。

Nowadays the one-way traffic is introduced all over the world in order to improve driving safety and reduce traffic jams. The government of Berland decided to keep up with new trends. Formerly all n cities of Berland 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 Berland 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

The first line contains integer n (3 ≤ n ≤ 100) — amount of cities (and roads) in Berland. 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 aito city bi, redirecting the traffic costs ci.

Output

Output single integer — 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 test(s)

input
3
1 3 1
1 2 1
3 2 1

output
1

input
3
1 3 1
1 2 5
3 2 1

output
2

input
6
1 5 4
5 3 8
2 4 15
1 6 16
2 3 23
4 6 42

output
39

input
4
1 2 9
2 3 8
3 4 7
4 1 5

output
0

我还是想问，这个不是建立联系表么……为啥这道题要算作是dfs啊……还是说这个就是dfs的意思么……

这道题是要求形成一个单向环（最初吓了我一跳，还以为是图论题，吾辈最怕图论了……），后来发现……只需要一条路找下去，不行的就花钱逆回来，然后看看正着便宜还是倒着便宜就行了。

#include <map>
#include <stack>
#include <queue>
#include <cmath> 
#include <cctype>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

int  n=0,pos=0,ans1=0,ans2=0;
bool vis	[111];
int  chain	[111];
int  cost	[111][111];

#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))

void dfs(int now)// make a linked chain of nodes
{
	vis[now]=true;
	chain[pos++]=now;
	for(int k=1;k<=n;k++)// find next
	{
		if(vis[k]==false && (cost[k][now]>0 || cost[now][k]>0))
		dfs(k);
	}
}

int main()
{
	memset(cost,0,sizeof(cost));
	memset(vis,false,sizeof(vis));
	int total_cost=0;
	scanf("%d",&n);
	for(int i=0;i<n;i++)
	{
		int _start=0,_end=0;
		scanf("%d%d",&_start,&_end);
		scanf("%d",&cost[_start][_end]);
		total_cost+=cost[_start][_end];
	}
	dfs(1);// start from 1,make a chain until 1 ends
	chain[pos]=1;// when end, we get back to 1
	for(int j=0;j<pos;j++)		ans1+= cost[chain[j]][chain[j+1]];
	for(int j=pos;j>=1;j--)		ans2+= cost[chain[j]][chain[j-1]];
	//cout<<total_cost<<endl<<ans1<<endl<<ans2<<endl;
	printf("%d",Min(ans1,ans2));
	return 0;
}