P1223 [小数据版]边权差值最小的生成树

　　这道题和最小生成树kruskal的代码几乎相同，只不过不一定是最小生成树，所以不一定从最短的边开始做生成树；所以将每一条边分别作为起点，然后枚举就行了......

　　

#include <bits/stdc++.h>
 
using namespace std;
 
#define INF 0x3f3f3f3f
 
inline int read()
{
    int x = ,ff = ;char ch = getchar();
    while(!isdigit(ch))
    {
        if(ch == '-') ff = -;
        ch = getchar();
    }
    while(isdigit(ch))
    {
        x = (x<<) + (x<<) + (ch ^ );
        ch = getchar();
    }
    return x * ff;
}
 
inline void write(int x)
{
    if(x < ) putchar('-'),x = -x;
    if(x > ) write(x / );
    putchar(x %  + '');
}
 
struct node
{
    int x,y,v;
}e[];
int a,b,x,y,v,m = ,fa[],mi = INF;
 
inline bool cmp(node x,node y)
{
    return x.v < y.v;
}
 
int get(int x)
{
    return (x == fa[x]) ? x : (fa[x] =  get(fa[x]));
}
 
int Kruskal(int ans)
{
    int maxx = -,minn = INF;
    for(int i = ;i <= a;++i)
    fa[i] = i;
    int sum = ,cnt = ;
    sort(e + ,e + m + ,cmp);
    for(int i = ans;cnt < a - ;++i)
    {
        ++sum;
        if(sum > m) return -; // 所有的边都找过，不能构成树；
        int x = get(e[i % m].x),y = get(e[i % m].y); // i可能大于m，所以用i % m；
        if(x != y)
        {
            ++cnt;
            fa[x] = y;
            maxx = max(maxx,e[i % m].v);
            minn = min(minn,e[i % m].v);
        }
    }
    return maxx - minn;
}
 
int main()
{
    a = read(); b = read();
    for(int i = ;i <= b;++i)
    {
        x = read(); y = read(); v = read();
        e[++m].x = x; e[m].y = y; e[m].v = v;
    }
    for(int i = ;i <= m;++i)
        mi = min(mi,Kruskal(i)); // 枚举不同的起点
    write(mi);
    return ;
}