Codeforces Gym 100803F There is No Alternative 暴力Kruskal

There is No Alternative

题目连接：

http://codeforces.com/gym/100803/attachments

Description

ICPC (Isles of Coral Park City) consist of several beautiful islands.

The citizens requested construction of bridges between islands to resolve inconveniences of using

boats between islands, and they demand that all the islands should be reachable from any other

islands via one or more bridges.

The city mayor selected a number of pairs of islands, and ordered a building company to estimate

the costs to build bridges between the pairs. With this estimate, the mayor has to decide the

set of bridges to build, minimizing the total construction cost.

However, it is difficult for him to select the most cost-efficient set of bridges among those

connecting all the islands. For example, three sets of bridges connect all the islands for the

Sample Input 1. The bridges in each set are expressed by bold edges in Figure F.1.

Figure F.1. Three sets of bridges connecting all the islands for Sample Input 1

As the first step, he decided to build only those bridges which are contained in all the sets

of bridges to connect all the islands and minimize the cost. We refer to such bridges as no

alternative bridges. In Figure F.2, no alternative bridges are drawn as thick edges for the

Sample Input 1, 2 and 3.

Write a program that advises the mayor which bridges are no alternative bridges for the given

input.

Input

The first line contains two positive integers N and M. N represents the number of islands and

each island is identified by an integer 1 through N. M represents the number of the pairs of

islands between which a bridge may be built.

Each line of the next M lines contains three integers Si

, Di and Ci (1 ≤ i ≤ M) which represent

that it will cost Ci to build the bridge between islands Si and Di

. You may assume 3 ≤ N ≤ 500,

N − 1 ≤ M ≤ min(50000, N(N − 1)/2), 1 ≤ Si < Di ≤ N, and 1 ≤ Ci ≤ 10000. No two bridges

connect the same pair of two islands, that is, if i 6= j and Si = Sj , then Di 6= Dj . If all the

candidate bridges are built, all the islands are reachable from any other islands via one or more

bridges.

Output

Output two integers, which mean the number of no alternative bridges and the sum of their

construction cost, separated by a space.

Sample Input

4 4

1 2 3

1 3 3

2 3 3

2 4 3

Sample Output

1 3

Hint

题意

给你一个图，然后问你有哪些边一定在所有的最小生成树上面

题解：

点只有500，所以我们直接O(nm)暴力就好了

我们只用ban掉一开始在最小生成树上的边，然后判断就好了

代码

#include<bits/stdc++.h>
using namespace std;

const int maxn = 100050;

int u[maxn],v[maxn],w[maxn],r[maxn];
int pa[maxn],n,m;
int vis[maxn];
int ans1=0,ans2=0;
bool cmp(int i,int j)
{
    return w[i]<w[j];
}
int fi(int x)
{
    return pa[x]==x?x:pa[x]=fi(pa[x]);
}
int kruskal(int c)
{
    int ans=0;
    int num=0;
    for(int i=0;i<=n;i++)
        pa[i]=i;
    for(int i=0;i<m;i++)
    {
        if(i==c)continue;
        int e=r[i];
        int x=fi(u[e]);
        int y=fi(v[e]);
        if(x!=y)
        {
            num++;
            vis[i]=1;
            ans+=w[e],pa[x]=y;
        }
        if(num==n-1)break;
    }
    if(num!=n-1)return -1;
    return ans;
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=0;i<m;i++)
    {
        int x,y,z;
        scanf("%d%d%d",&x,&y,&z);
        u[i]=x,v[i]=y,w[i]=z;
    }
    for(int i=0;i<m;i++)
        r[i]=i;
    sort(r,r+m,cmp);
    int temp = kruskal(m);
    for(int i=0;i<m;i++)
    {
        if(!vis[i])continue;
        if(temp!=kruskal(i))
        {
            ans1++;
            ans2+=w[r[i]];
        }
    }
    cout<<ans1<<" "<<ans2<<endl;
}