POJ-2377 Bad Cowtractors

题目传送门
Bad Cowtractors
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 16504 Accepted: 6714
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.
Source

USACO 2004 December Silver

这个题是最小生成树的变形题，求最大生成树，这个也好办，最小生成树求的是边权和最小的一棵树，那么我们把每个正边权取为负边权，再去求最小生成树，得到的最小的一棵树的边权和，肯定是一个负数，并且是最小的，那么它的相反数就是最大的正边权和的一棵树，就是最大生成树。

具体看代码，我习惯用prim。

代码：

#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#include<vector>#include<cmath>using namespace std;typedef pair<int,int>P;const int maxn=1000+10;const int INF=0x3f3f3f3f;struct Edge{    int to,len;    Edge(int to_=0,int len_=0)    {        to=to_;        len=len_;    }};vector<Edge>G[maxn];int dis[maxn];bool vis[maxn];int n,m;void init(){    for(int i=1;i<=n;i++)    G[i].clear();}void Add(int from,int to,int len){    G[from].push_back(Edge(to,len));    G[to].push_back(Edge(from,len));}int prime(int s){    fill(dis,dis+n+1,INF);    dis[s]=0;    memset(vis,false,sizeof(vis));    priority_queue<P,vector<P>,greater<P> >que;    int u,v,i,len,cnt=0,ans=0;    Edge e;    P p;    que.push(P(0,s));    while(!que.empty())    {        p=que.top();        que.pop();        u=p.second;        if(vis[u])        continue;        vis[u]=true;        cnt++;        ans+=dis[u];        len=G[u].size();        for(int i=0;i<len;i++)        {            e=G[u][i];            v=e.to;            if(!vis[v]&&dis[v]>e.len)            {                 dis[v]=e.len;                 que.push(P(dis[v],v));             }         }    }    if(cnt!=n)    return -1;    return -ans;}int main(){    while(scanf("%d%d",&n,&m)==2)    {        init();        for(int i=0;i<m;i++)        {           int u,v,len;           scanf("%d%d%d",&u,&v,&len);           Add(u,v,-len);        }        printf("%d\n",prime(1));    }}