Bond UVA

题意：

　　给你一张无向图，然后有若干组询问，让你输出a->b的最小瓶颈路。

解析：

　　应该都想过用prime的次小生成树做。。但二维数组开不了那么大。。所以只能用kruskal了。。。。

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _  ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = 50010, INF = 0x7fffffff, maxm = 1e5+10;
int n, m;
int f[maxn], ra[maxn], vis[maxn], edge[maxn];

struct node
{
    int u, v, w;
}Node[maxm];

bool cmp(node a, node b)
{
    return a.w < b.w;
}
int find(int x)
{
    return f[x] == x ? x : find(f[x]);
}

int query(int x, int y)
{
    int ans1 = 0, ans2 = -1;
    int cur = x;
    while(1)  //从x回溯到祖先
    {
        vis[cur] = ans1;  //标记 从x到当前cur的最大权值的路段
        if(cur == f[cur]) break;
        ans1 = max(ans1, edge[cur]);
        cur = f[cur];
    }
    cur = y;
    while(1)  //从y回溯到祖先
    {
        if(vis[cur] >= 0)  //直到遇到y和x的最近公共祖先
        {
            ans2 = max(ans2, vis[cur]);
            break;
        }
        ans2 = max(ans2, edge[cur]);
        cur = f[cur];
    }
    cur = x;
    while(1)   //还原vis。。其实用一个memset就好了。。但时间复杂度竟然比用这个大10倍。。emm。。。
    {
        vis[cur] = -1;
        if(cur == f[cur]) break;
        cur = f[cur];
    }
    return ans2;
}

void init()
{
    rap(i, 1, n)
    {
        f[i] = i;
        ra[i] =1;
    }
    mem(vis, -1);
}

int main()
{
    bool flag = true;
    while(~scanf("%d%d", &n, &m))
    {
        init();
        rap(i, 1, m)
        {
            scanf("%d%d%d", &Node[i].u, &Node[i].v, &Node[i].w);
        }
        sort(Node+1, Node+m+1, cmp);
        rap(i, 1, m)  //按秩合并
        {
            int l = find(Node[i].u);
            int r = find(Node[i].v);
            if(l == r) continue;
            if(ra[l] <= ra[r]) f[l] = r, ra[r] = max(ra[r], ra[l] + 1), edge[l] = Node[i].w;  //儿子标记权值
            else f[r] = l, ra[l] = max(ra[l], ra[r] + 1), edge[r] = Node[i].w;
        }
        if(flag) flag = false;
        else printf("
");
        int q;
        scanf("%d", &q);
        rap(i, 1, q)
        {
            int u, v;
            scanf("%d%d", &u, &v);
            printf("%d
", query(u, v));
        }

    }

    return 0;
}