Out of Hay

来源：USACO 2005 March Silver 

OJ题号：洛谷1547、POJ2395

思路：

Kruskal求最小生成树，记录最大边。

#include<cstdio>
#include<cctype>
#include<algorithm>
inline int getint() {
    char ch;
    while(!isdigit(ch=getchar()));
    int x=ch^'0';
    while(isdigit(ch=getchar())) x=(((x<<2)+x)<<1)+(ch^'0');
    return x;
}
class DisjointSet {
    private:
        static const int N=2001;
        int anc[N];
        int Find(const int x) {
            return (x==anc[x])?x:(anc[x]=Find(anc[x]));
        }
    public:
        DisjointSet(const int n) {
            for(int i=1;i<=n;i++) {
                anc[i]=i;
            }
        }
        bool isConnected(const int x,const int y) {
            return Find(x)==Find(y);
        }
        void Union(const int x,const int y) {
            anc[Find(x)]=Find(y);
        }
};
struct Edge {
    int u,v,w;
    bool operator < (const Edge &x) const {
        return w<x.w;
    }
};
int main() {
    int n=getint(),m=getint();
    DisjointSet s(n);
    Edge e[m];
    for(int i=0;i<m;i++) {
        e[i].u=getint(),e[i].v=getint(),e[i].w=getint();
    }
    std::sort(&e[0],&e[m]);
    int ans=0;
    for(int i=0;i<m;i++) {
        if(s.isConnected(e[i].u,e[i].v)) continue;
        s.Union(e[i].u,e[i].v);
        ans=e[i].w;
    }
    printf("%d\n",ans);
    return 0;
}