【UVA 11354】 Bond （最小瓶颈生成树、树上倍增）

【题意】

　　n个点m条边的图 q次询问 找到一条从s到t的一条边 使所有边的最大危险系数最小

Input
There will be at most 5 cases in the input file.
The first line of each case contains two integers N, M (2 ≤ N ≤ 50000, 1 ≤ M ≤ 100000) – number
of cities and roads. The next M lines describe the roads. The i-th of these lines contains three integers:
xi, yi, di (1 ≤ xi, yi ≤ N, 0 ≤ di ≤ 10^9) – the numbers of the cities connected by the i-th road and its
dangerousness.
Description of the roads is followed by a line containing an integer Q (1 ≤ Q ≤ 50000), followed by
Q lines, the i-th of which contains two integers si and ti (1 ≤ si
, ti ≤ N, si ̸= ti).
Consecutive input sets are separated by a blank line.
Output
For each case, output Q lines, the i-th of which contains the minimum dangerousness of a path between
cities si and ti
. Consecutive output blocks are separated by a blank line.
The input file will be such that there will always be at least one valid path.
Sample Input
4 5
1 2 10
1 3 20
1 4 100
2 4 30
3 4 10
2
1 4
4 1
2 1
1 2 100
1
1 2
Sample Output
20
20
100

【分析】

　　很明显是最小瓶颈生成树。

　　有一个定理：最小生成树是最小瓶颈生成树，但是最小瓶颈生成树不一定是最小生成树。

　　我们只要求最小生成树就好了。

　　不过这题n较大，不能n^2预处理，所以我们先把树求出来，然后询问的时候 树剖或者倍增 都可以。

　　应该是，倍增耗空间但是时间少一个log ， 树剖省一点空间但是 时间多一个log （要多一个数据结构维护）

　　我上次就打一题倍增MLE了TAT，不过这题正解是倍增，不知道树剖+线段树能不能过。

 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #include<iostream>
 #include<algorithm>
 #include<queue>
 using namespace std;
 #define Maxn 50010
 #define Maxm 1000010
 #define INF 0xfffffff

 struct node
 {
     int x,y,c,next;
 }t[Maxn*],tt[Maxm];
 int len;
 int first[Maxn];

 bool cmp(node x,node y) {return x.c<y.c;}
 int mymax(int x,int y) {return x>y?x:y;}

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

 void ins(int x,int y,int c)
 {
     t[++len].x=x;t[len].y=y;t[len].c=c;
     t[len].next=first[x];first[x]=len;
 }

 int f[Maxn][],g[Maxn][],dep[Maxn];

 void dfs(int x,int ff,int l)
 {
     dep[x]=dep[ff]+;
     g[x][]=ff;
     for(int i=;(<<i)<=dep[x];i++)
        g[x][i]=g[g[x][i-]][i-];
     f[x][]=l;
     for(int i=;(<<i)<=dep[x];i++)
       f[x][i]=mymax(f[x][i-],f[g[x][i-]][i-]);
     for(int i=first[x];i;i=t[i].next) if(t[i].y!=ff)
         dfs(t[i].y,x,t[i].c);
 }

 int ffind(int x,int y)
 {
     int ans=;
     while(dep[x]!=dep[y])
     {
         int z;
         if(dep[x]<dep[y]) z=x,x=y,y=z;
         for(int i=;i>=;i--) if(dep[x]-(<<i)>=dep[y])
             ans=mymax(ans,f[x][i]),x=g[x][i];
     }
     if(x==y) return ans;
     if(x!=y)
     {
         for(int i=;i>=;i--) if(g[x][i]!=g[y][i]&&dep[x]>=(<<i))
             ans=mymax(ans,f[x][i]),ans=mymax(ans,f[y][i]),
             x=g[x][i],y=g[y][i];
     }
     ans=mymax(ans,f[x][]);ans=mymax(ans,f[y][]);
     return ans;
 }

 int main()
 {
     int n,m;
     bool ok=;
     while(scanf("%d%d",&n,&m)!=EOF)
     {
         if(ok) printf("\n");
         ok=;
         for(int i=;i<=m;i++)
         {
             scanf("%d%d%d",&tt[i].x,&tt[i].y,&tt[i].c);
         }
         sort(tt+,tt++m,cmp);
         int cnt=;
         for(int i=;i<=n;i++) fa[i]=i;
         len=;
         memset(first,,sizeof(first));
         for(int i=;i<=m;i++)
         {
             if(ffind(tt[i].x)!=ffind(tt[i].y))
             {
                 fa[ffind(tt[i].x)]=ffind(tt[i].y);
                 cnt++;
                 ins(tt[i].x,tt[i].y,tt[i].c);
                 ins(tt[i].y,tt[i].x,tt[i].c);
             }
             if(cnt==n-) break;
         }
         dep[]=;
         dfs(,,);
         int q;
         scanf("%d",&q);
         for(int i=;i<=q;i++)
         {
             int x,y;
             scanf("%d%d",&x,&y);
             printf("%d\n",ffind(x,y));
         }
     }
     return ;
 }

2016-11-01 08:16:45