uva 11354 Bond

题目：Bond

题意：给出一张无向图和m个询问x y，输出x，y之间的最小瓶颈路。

思路：用lca倍增求最小瓶颈路。

注意：当x和y的lca是y时要特判。

代码：

#include<bits/stdc++.h>
using namespace std;

#define maxn 50000
#define maxm 100000
#define maxl 20

struct Edge {
	int x,y,z;
	Edge(int xx=0,int yy=0,int zz=0) {
		x=xx,y=yy,z=zz;
	}
	bool operator < (const Edge& oth) const {
		return z<oth.z||(z==oth.z&&(x<oth.x||(x==oth.x&&y<oth.y)));
	}
};

struct Node {
	int fa,c,d;
};

int n,m;
Edge a[maxm+5];
int fa[maxm+5];

vector<Edge> tr[maxn+5];

Node e[maxn+5];

int anc[maxn+5][maxl+5];
int mc[maxn+5][maxl+5];

void init() {
	for(int i=1; i<=maxm; i++) fa[i]=i;
	for(int i=1; i<=maxn; i++) tr[i].clear();
	memset(mc,0,sizeof(mc));
	memset(anc,0,sizeof(anc));
}

void readin() {
	for(int i=1; i<=m; i++) {
		scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].z);
	}
}

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

int kruskal() {
	int ans=0,cnt=0;
	for(int i=1; i<=m; i++) {
		int fa1=find(a[i].x),fa2=find(a[i].y);
		if(fa1==fa2) continue;
		fa[fa1]=fa2;
		ans+=a[i].z;
		cnt++;

		tr[a[i].x].push_back(a[i]);
		tr[a[i].y].push_back(Edge(a[i].y,a[i].x,a[i].z));

		if(cnt==n-1) break;
	}
	return ans;
}

void chg(int x,int fa,int d) {
	e[x].fa=fa,e[x].d=d;
	for(int i=0; i<tr[x].size(); i++) {
		int y=tr[x][i].y;
		if(y!=fa) e[y].c=tr[x][i].z,chg(y,x,d+1);
	}
	return ;
}

int query(int x,int y) {
	int p;
	for(p=1; (1<<p)<=e[x].d; p++);
	p--;

	int ans=-(1<<30);
	for(int i=p; i>=0; i--) {
		if(e[x].d>=(1<<i)+e[y].d) {
			ans=max(ans,mc[x][i]);
			x=anc[x][i];
		}
	}
	if(x==y) return ans;
	
	for(int i=p; i>=0; i--) {
		if(anc[x][i]&&anc[x][i]!=anc[y][i]) {
			ans=max(ans,mc[x][i]),ans=max(ans,mc[y][i]);
			x=anc[x][i],y=anc[y][i];
		}
	}
	ans=max(ans,e[x].c),ans=max(ans,e[y].c);
	
	return ans;
}

void slv() {
	int q;
	scanf("%d",&q);
	while(q--) {
		int x,y;
		scanf("%d%d",&x,&y);
		if(e[x].d<e[y].d) swap(x,y);
		printf("%d\n",query(x,y));
	}
}

void mk() {
	for(int i=1; i<=n; i++) {
		anc[i][0]=e[i].fa;
		mc[i][0]=e[i].c;
	}
	for(int j=1; (1<<j)<n; j++) {
		for(int i=1; i<=n; i++) {
			if(anc[i][j-1]) {
				anc[i][j]=anc[anc[i][j-1]][j-1];
				mc[i][j]=max(mc[i][j-1],mc[anc[i][j-1]][j-1]);
			}
		}
	}
}

int main() {
	int T=0;
	while(~scanf("%d%d",&n,&m)) {
		if(T) printf("\n");
		else T=1;
		init();
		readin();
		sort(a+1,a+m+1);
		int s=kruskal();
		chg(1,1,0);
		mk();
		slv();
	}
	return 0;
}