## Output

对每个询问，输出最长的边最小值是多少。

6 6 8
1 2 5
2 3 4
3 4 3
1 4 8
2 5 7
4 6 2
1 2
1 3
1 4
2 3
2 4
5 1
6 2
6 1

5
5
5
4
4
7
4
5

## HINT

1 <= N <= 15,000

1 <= M <= 30,000

1 <= d_j <= 1,000,000,000

1 <= K <= 15,000

1.构造最小生成树（无根树）。

2.将最小生成树构造为有根数，并用倍增LCA求出每个节点到第2^i个祖先的路径上的最长边。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
#define ms(a,b) memset((a),(b),sizeof((a)))
using namespace std;
typedef long long LL;
const int INF = 2e9;
const LL LNF = 9e18;
const int mod = 1e9+7;
const int maxn = 3e4+10;
const int DEG = 20;

struct node
{
int u, v, w;
bool operator<(const node &x)const {
return w<x.w;
}
}e[maxn];

struct edge
{
int to, w, next;
}edge[maxn*2];

int n, m,k;
int fa[maxn][DEG], deg[maxn], ma[maxn][DEG], be[maxn];

int find(int x) { return be[x]==x?x:x=find(be[x]); }

void add(int u, int v, int w)
{
edge[tot].to = v;
edge[tot].w = w;
}

void bfs(int root)
{
queue<int>que;
deg[root] = 0;
ma[root][0] = 0;
fa[root][0] = root;
que.push(root);
while(!que.empty())
{
int tmp = que.front();
que.pop();
for(int i = 1; i<DEG; i++)
fa[tmp][i] = fa[fa[tmp][i-1]][i-1], ma[tmp][i] = max( ma[tmp][i-1], ma[fa[tmp][i-1]][i-1]);
for(int i = head[tmp]; i!=-1; i = edge[i].next)
{
int v = edge[i].to, w = edge[i].w;
if(v==fa[tmp][0]) continue;
deg[v] = deg[tmp]+1;
fa[v][0] = tmp;
ma[v][0] = w;
que.push(v);
}
}
}

int LCA(int u, int v)
{
int ans = 0;
if(deg[u]>deg[v]) swap(u,v);
int hu = deg[u], hv = deg[v];
int tu = u, tv = v;
for(int det = hv-hu, i = 0; det; det>>=1, i++)
if(det&1)
ans = max(ans, ma[tv][i]), tv = fa[tv][i];

if(tv==tu) return ans;
for(int i = DEG-1; i>=0; i--)
{
if(fa[tu][i]==fa[tv][i]) continue;
ans = max(ans, max( ma[tu][i], ma[tv][i] ) );
tu = fa[tu][i];
tv = fa[tv][i];
}
return ans = max(ans, max( ma[tu][0], ma[tv][0]) );
}

int main()
{
tot = 0;
ms(ma,0);
scanf("%d%d%d",&n,&m,&k);
for(int i = 0; i<m; i++)
scanf("%d%d%d",&e[i].u,&e[i].v,&e[i].w);

sort(e,e+m);
for(int i = 1; i<=n; i++)
be[i] = i;
for(int i = 0; i<m; i++)
{
int u = find(e[i].u), v = find(e[i].v);
if(u!=v)
{
be[u] = v;
}
}

bfs(1);
for(int i = 0; i<k; i++)
{
int u, v;
scanf("%d%d",&u,&v);
printf("%d\n",LCA(u,v));
}
}


posted on 2017-07-21 11:03  h_z_cong  阅读(154)  评论(0编辑  收藏  举报