HDU 1598 find the most comfortable road 並查集+貪心

對邊做一個排序，然後對於每一個詢問，每次都去枚舉權值最小的邊，再做最小生成樹，一旦詢問中的點聯通了，我們就可以認爲在當前最小邊權值的情況下，舒適度最大的情況已經達到了。

#include<iostream>
#include<cstdio>
#include<string.h>
#include<algorithm>
#include<cmath>
using namespace std;

int fa[210],n,m,q;

struct node{
    int x,y,w;
    bool operator < (const node &n) const {return w<n.w;}
}edge[1010];

int find(int x)
{
    if (fa[x]!=x) fa[x]=find(fa[x]);
    return fa[x];
}

int join(int a,int b)
{
    int ret=0;
    int fx=find(a); int fy=find(b);
    if (fx!=fy)
    {
        fa[fx]=fy;
        ret=1;
    }
    return ret;
}

int main()
{
    while (~scanf("%d%d",&n,&m))
    {
        for (int i=1; i<=m; i++)
        {
            scanf("%d%d%d",&edge[i].x,&edge[i].y,&edge[i].w);
        }
        sort(edge+1,edge+1+m);
        scanf("%d",&q);
        for (int i=1; i<=q; i++)
        {
            int a,b; scanf("%d%d",&a,&b);
            int ans=1000000;
            for (int j=1; j<=m; j++)
            {
                //cout<<"j="<<edge[j].w<<endl;
                int tmp=-1;
                for (int k=1; k<=n; k++)
                {
                    fa[k]=k;
                }
                for (int k=j; k<=m; k++)
                {
                    //cout<<"k="<<edge[k].w<<endl;
                    join(edge[k].x,edge[k].y);
                    //cout<<"1="<<edge[k].x<<" 2="<<edge[k].y<<endl;
                    if (find(a)==find(b))
                    {
                        tmp=edge[k].w-edge[j].w;
                        //cout<<"tmp="<<tmp<<endl;
                        break;
                    }
                }
                if (tmp!=-1) ans=min(ans,tmp);
            }
            if (ans==1000000) printf("-1\n");
            else printf("%d\n",ans);
        }
    }
    return 0;
}