POJ 2377 Bad Cowtractors

Bad Cowtractors

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12441		Accepted: 5168

Description

Bessie has been hired to build a cheap internet network among Farmer John's N (2 <= N <= 1,000) barns that are conveniently numbered 1..N. FJ has already done some surveying, and found M (1 <= M <= 20,000) possible connection routes between pairs of barns. Each possible connection route has an associated cost C (1 <= C <= 100,000). Farmer John wants to spend the least amount on connecting the network; he doesn't even want to pay Bessie. 

Realizing Farmer John will not pay her, Bessie decides to do the worst job possible. She must decide on a set of connections to install so that (i) the total cost of these connections is as large as possible, (ii) all the barns are connected together (so that it is possible to reach any barn from any other barn via a path of installed connections), and (iii) so that there are no cycles among the connections (which Farmer John would easily be able to detect). Conditions (ii) and (iii) ensure that the final set of connections will look like a "tree".

Input

* Line 1: Two space-separated integers: N and M 

* Lines 2..M+1: Each line contains three space-separated integers A, B, and C that describe a connection route between barns A and B of cost C.

Output

* Line 1: A single integer, containing the price of the most expensive tree connecting all the barns. If it is not possible to connect all the barns, output -1.

Sample Input

5 8
1 2 3
1 3 7
2 3 10
2 4 4
2 5 8
3 4 6
3 5 2
4 5 17

Sample Output

42

Hint

OUTPUT DETAILS: 

The most expensive tree has cost 17 + 8 + 10 + 7 = 42. It uses the following connections: 4 to 5, 2 to 5, 2 to 3, and 1 to 3.

题意：贝西承包了联接n个农场间光纤的工程，现在给出m条路线，a  b  c  表示从a农场到b农场连通光纤要c元。 贝西想多赚一点钱，所以尽量选择花费多的路线联接，要保证连接上所有的农场，但是由不能形成环路，那样农场主约翰会知道。 问约翰最多要花费多少钱，不能连通输出-1。

很裸的最大生成树 但是要注意重编问题

用最小生成树Kruskal的做法，但按照大边排序，因为每次要取最大的，才能保证最后形成的树cost最大，不要忘记图不连通的情况，应该用并查集合并操作的次数来判定，当合并操作的次数为顶点数减一时，图就是连通的

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
struct edge{
	ll u,v,cost;
}node[21000];
ll n,m,pre[1100];
bool cmp(edge a,edge b){
	return a.cost>b.cost;
}
ll find(ll x){
	if (pre[x]==x) return x;
	else
	return pre[x]=find(pre[x]);
}
void init(ll n)
{
   	for (int i=1;i<=n;i++)
		pre[i]=i;
}
void mix(int a,int b)
{
    int fx=find(a);
    int fy=find(b);
    if(fx!=fy)
    {
        pre[fx]=fy;
    }
}
int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<=m;i++)
		scanf("%d%d%d",&node[i].u,&node[i].v,&node[i].cost);
	sort(node+1,node+1+m,cmp);
    init(n);
	ll cnt=0,ans=0;
	for (int i=1;cnt<n-1&&i<=m;i++){
		if (find(node[i].u)!=find(node[i].v))
        {
            ans+=node[i].cost;
            cnt++;
            mix(node[i].u,node[i].v);
        }
	}
	if (cnt==n-1)
		printf("%lld",ans);
	else printf("-1");
	return 0;
}

prim算法

#include<cstdio>
#include<cstring>
#define maxn 2020
#define INF 0x3f3f3f3f
using namespace std;
int map[maxn][maxn],n;
 
void prim()
{
	int dis[maxn],visit[maxn];
	int i,j,min,next,maxlen=0;
	memset(visit,0,sizeof(visit));
	for(i=1;i<=n;++i)
		dis[i]=map[1][i];
	visit[1]=1;
	for(i=2;i<=n;++i)
	{
		min=INF;
		for(j=1;j<=n;++j)
		{
			if(!visit[j]&&dis[j]<min)
			{
				min=dis[j];
				next=j;
			}
		}
		if(min!=INF&&min>maxlen)
			maxlen=min;
		visit[next]=1;
		for(j=1;j<=n;++j)
		{
			if(!visit[j]&&dis[j]>map[next][j])
				dis[j]=map[next][j];
		}
	}
	printf("%d\n",maxlen);
}
 
int main()
{
	int m,i,j,a,b,c;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(i=1;i<=n;++i)
		{
			for(j=i;j<=n;++j)
			{
				if(i==j)
					map[i][j]=0;
				else
					map[i][j]=map[j][i]=INF;
			}
		}
		while(m--)
		{
			scanf("%d%d%d",&a,&b,&c);
			if(map[a][b]>c)
				map[a][b]=map[b][a]=c;
		}
		prim();
	}
	return 0;
}