POJ2377 解题报告

Bad Cowtractors

Time Limit: 1000MSMemory Limit: 65536KTotal Submissions: 15459Accepted: 6344

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 81 2 31 3 72 3 102 4 42 5 83 4 63 5 24 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

模板题，由于最后要判断还不是一棵树，所以我选择使用kruskal算法，因为里面用到了并查集，通过并查集我可以很快判断出所有的节点是否连在一棵树上

并且这题问的是最大的cost，那我就是需要把我的edge按照e.cost从大到小排列就好了

总之直接套模板，改动不大

//POJ2377#include<iostream>#include<algorithm>using namespace std;struct edge{    int u;    int v;    int cost;};const int INF=0x7fffffff;const int maxm=20000+20; //边数最大值const int maxn=1000+100; //节点最大值const int maxc=100000; //cost最大值edge es[maxm];int par[maxn];int rnk[maxn];int N,M;bool cmp(const edge  &e1,const edge &e2);int kruskal();void init(int n);void unite(int x,int y);int fin(int x);bool same(int x,int y);int main(){    cin>>N>>M;    for(int i=0;i<M;i++)    {        int a,b,c;        cin>>a>>b>>c;        es[i].u=a; es[i].v=b; es[i].cost=c;    }    cout<<kruskal()<<endl;    return 0;}bool cmp(const edge  &e1,const edge &e2){    return e1.cost>e2.cost;}int kruskal(){    init(N);    sort(es,es+M,cmp);    int ans=0;    for(int i=0;i<M;i++)    {        edge e=es[i];        if(!same(e.u,e.v))        {            unite(e.u,e.v);            ans+=e.cost;        }    }    int par1=fin(1);    for(int i=2;i<=N;i++)        if(par1!=fin(i)) return -1;    return ans;}void init(int n){    for(int i=1;i<=N;i++)    {        par[i]=i;        rnk[i]=0;    }}void unite(int x,int y){    x=fin(x);    y=fin(y);    if(x==y) return;    if(rnk[x]<rnk[y]) par[x]=y;    else    {        par[y]=x;        if(rnk[x]==rnk[y]) rnk[x]++;    }}int fin(int x){    if(x==par[x]) return x;    return par[x]=fin(par[x]);}bool same(int x,int y){    return fin(x)==fin(y);}