poj 1251 Jungle Roads(最小生成树)

题目链接：http://poj.org/problem?id=1251

思路分析：使用最小生成树算法，可以求解。需要注意的树Kruskal算法中使用了并查集，对于并查集用法需要注意。

 

代码如下： 

#include <iostream>
#include <algorithm>
using namespace std;

const int N = 30, M = 100;
int S[N];
int Index, n;
typedef struct { int U, V, W; }Edge;
Edge E[M];

bool cmp( Edge a, Edge b ) { return a.W < b.W; }
int Find( int X )
{
    if ( S[X] <= 0 )
        return X;
    else
        return S[X] = Find( S[X] );
}

void SetUnion( int Root1, int Root2 )
{
    int Rank_Root1 = Find( Root1 );
    int Rank_Root2 = Find( Root2 );
    int Tmp = S[Rank_Root1] + S[Rank_Root2];
   
    if ( S[Rank_Root1] < S[Rank_Root2] )
    {
        S[Rank_Root2] = Rank_Root1;
        S[Rank_Root1] = Tmp;
    }
    else
    {
        S[Rank_Root1] = Rank_Root2;
        S[Rank_Root2] = Tmp;
    }
}


int Kruskal()
{
    int Ans = 0, Cnt = 0;

    memset( S, -1, sizeof(S) );
    for ( int i = 0; i < Index; i++ )
    {
        int U = E[i].U;
        int V = E[i].V;
        int SetU = Find( U );
        int SetV = Find( V ); 
        if ( Find(U) != Find(V) )
        {
            Cnt++;
            Ans += E[i].W;
            SetUnion( U, V );
        }if ( Cnt >= n - 1 )
            break;
    }

    return Ans;
}

int main()
{
    int w;
    int Ans, Num;
    char u, v;

    while ( cin >> n )
    {
        if ( n == 0 )
            break;

        Index = 0;
        for ( int i = 0; i < n-1; ++i )
        {
            cin >> u >> Num;
            for ( int j = 0; j < Num; ++j )
            {
                cin >> v >> w;
                E[Index].U = u - 'A';
                E[Index].V = v - 'A';
                E[Index].W = w;
                Index++;
            }
        }

        sort( E, E+Index, cmp );
        Ans = Kruskal();
        cout << Ans << endl;
    }

    return 0;
}