Uva 1395

題目鏈接：

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=105277#problem/B

題意：

給出一個n節點的圖，邊最多爲n*(n-1)/2條邊，求苗條度（最大邊權值減最小邊權值）儘量小的生成樹。

分析：

按權值從小到大排序，從小到大枚舉L,然後直到連通邊數有n-1時，就枚舉另一個L，保存最優值就可以了。

#include <stdio.h>
#include <algorithm>
#define INF 0x7fffffff
const int maxn = 105;
struct node
{
    int x, y, v;
    friend bool operator < ( node n1, node n2 )
    {
        return n1.v < n2.v; //權值排序
    }
}a[maxn*maxn];
int fa[maxn];
void init ( int n )
{
    for ( int i = 1; i <= n; i ++ )
        fa[i] = i;
}
int find ( int x ) { return fa[x] == x ? x : fa[x] = find ( fa[x] ); }
inline int Min ( int a, int b ) { return a < b ? a : b; }
inline int Max ( int a, int b ) { return a > b ? a : b; }
int main ( )
{
    int n, m;
    while ( ~ scanf ( "%d%d", &n, &m ) && ( n || m ) )
    {
        for ( int i = 0; i < m; i ++ )
            scanf ( "%d%d%d", &a[i].x, &a[i].y, &a[i].v );
        std :: sort ( a, a+m );
        int ans = INF;
        for ( int i = 0; i < m; i ++ )
        {
            init ( n );
            int cnt = 0, mn, mx, j;
            mn = INF, mx = 0;
            for ( j = i; j < m; j ++ )
            {
                int fx = find ( a[j].x ), fy = find ( a[j].y );
                if ( fx != fy )
                {
                    fa[fx] = fy;
                    cnt ++; //統計連通的邊數
                    mn = Min ( mn, a[j].v );
                    mx = Max ( mx, a[j].v );
                }
                if ( cnt >= n-1 )   //n-1條證明生成樹已完成
                    break ;
            }
            if ( j < m )
                ans = Min ( ans, mx-mn );   //找最小值
        }
        printf ( "%d\n", ans == INF ? -1 : ans );
    }
    return 0;
}