csu 1930 roads(DFS)

Description

Once upon a time there was a strange kingdom, the kingdom had n cities which were connected by n directed roads and no isolated city.One day the king suddenly found that he can't get to some cities from some cities.How amazing!The king is petty so he won't build some new roads to improve this situation,but he has superpowers that he can change the direction of any road.To do this,he will gain a certain fatigue value for a certain road.The king didn't want to be too tired.So he want to know what is the smallest amount of fatigue value he will gain on the redirecting of roads so that from every city people can get to any other?

Input

The first line contains integer n (3<=n<=100) - amount of cities (and roads) in the king. 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 it costs ci.

Output

Output single integer - the smallest amount of fatigue value the king will gain on the redirecting of roads so that from every city people can get to any other.

Sample Input

3
1 3 1
1 2 1
3 2 1
3
1 3 1
1 2 5
3 2 1

Sample Output

1
2

看上去很難,稍加分析可知n個點n條邊改變方向後能夠連通,只有多是一個環,因此咱們判斷反向邊和正向邊分別的權值總和取個小的就能夠了.而後若是咱們對每條單向邊建一條負權值的反向邊，跑一遍DFS就能夠了.

#include <iostream>
#include <algorithm>
#include <string.h>
#include <stdio.h>
using namespace std;
const int N = 105;
int mp[N][N];
int vis[N];
int cost;
int n,node;
void dfs(int u,int pre){
    if(vis[u]==1&&u!=1){
        return;
    }
    vis[u]++;
    if(vis[u]==2&&u==1){
        node = pre;
        return;
    }
    for(int i=1;i<=n;i++){
        if(i==pre) continue;
        if(mp[u][i]&&!vis[i]){
            if(mp[u][i]<0) cost+=mp[u][i];
            dfs(i,u);
        }
        if(mp[u][i]&&vis[i]!=2&&i==1){
            if(mp[u][i]<0) cost+=mp[u][i];
            dfs(i,u);
        }
    }
}
int main()
{

    while(scanf("%d",&n)!=EOF){
        cost = 0;
        int sum = 0;
        memset(mp,0,sizeof(mp));
        memset(vis,0,sizeof(vis));
        for(int i=1;i<=n;i++){
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            mp[u][v] = w;
            sum+=w;
            mp[v][u] = -w;
        }
        dfs(1,-1);
        cost=-cost;
        printf("%d\n",min(sum-cost,cost));
    }
    return 0;
}