【DFS+题意】#24 A. Ring road

A. Ring road

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

31 3 11 2 13 2 1

output

1

input

31 3 11 2 53 2 1

output

2

input

61 5 45 3 82 4 151 6 162 3 234 6 42

output

39

input

41 2 92 3 83 4 74 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 endschain[pos]=1;// when end, we get back to 1for(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;}