CF_Edu.#51_Div.2_1051F_The Shortest Statement

F. The Shortest Statement

time limit per test:4 seconds
memory limit per test:256 megabytes
input:standard input
output:standard output

You are given a weighed undirected connected graph, consisting of $n$ vertices and $m$ edges.

You should answer $q$ queries, the i-th query is to find the shortest distance between vertices $u_i$ and $v_i$.

给定n个点，m条边，q个询问，求每次询问的两点间的最短路。

Input

The first line contains two integers $n$ and $m (1≤n,m≤10^5,m−n≤20)$ — the number of vertices and edges in the graph.

Next $m$ lines contain the edges: the i-th edge is a triple of integers $v_i,u_i,d_i (1≤u_i,v_i≤n,1≤d_i≤10^9,u_i≠v_i)$. This triple means that there is an edge between vertices ui and vi of weight $d_i$. It is guaranteed that graph contains no self-loops and multiple edges.

The next line contains a single integer $q (1≤q≤10^5)$ — the number of queries.

Each of the next $q$ lines contains two integers $u_i$ and $v_i (1≤u_i,v_i≤n)$ — descriptions of the queries.

Pay attention to the restriction $m−n ≤ 20$.

Output

Print$q$lines.

The i-th line should contain the answer to the i-th query — the shortest distance between vertices $u_i$ and $v_i$.

Examples

Input

3 3

1 2 3

2 3 1

3 1 5

3

1 2

1 3

2 3

Output

3

4

1

Input

8 13

1 2 4

2 3 6

3 4 1

4 5 12

5 6 3

6 7 8

7 8 7

1 4 1

1 8 3

2 6 9

2 7 1

4 6 3

6 8 2

8

1 5

1 7

2 3

2 8

3 7

3 4

6 8

7 8

Output

7

5

6

7

7

1

2

7

SOLUTION

本题一开始说是要求$n=10^5$的多源最短路我被吓到了，裸做单源最短路是不可能的，就又看到了边数的限制：$m-n\leq 20$。

这就意味着，在n，m很大的绝大多数情况下，20并不能造成很大的影响。

所以换而言之，这题的模型可以近似地看作是一棵树。因为对于绝大多数的点来说，它们要走的最短路径的确全是树上路径。

树上最短路？求LCA啊。这样我们就可以解决绝大多数的点。

不过那20条边的确不能忽视，因为可能存在更优解要经过那21条边中的一些。而那21条边影响的是什么呢？当然是每条边的两个端点分别关于其他点的最短路啊（存这种端点的时候一定记得要去重！！！）。因为本题的余边只有21条，就可以考虑暴力做至多42遍dijkstra。。。得到了关于以编号为$j$的点为起点，图上点$i$的单源最短路数组$dp[i][j]$。

所以到了最后，我们可以得出，在本题内，最短路的答案要么只从树上的LCA算得，要么就可能走那多余的21条边。

这题是根据数据性质来猜正解的好题。

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cstdlib>

#include <algorithm>

#include <queue>

using namespace std;

typedef long long LL;

#define Min(a,b) ((a<b)?a:b)

#define Max(a,b) ((a>b)?a:b)

const int N=101000,MN=45;

inline int read(){

	int x=0,f=1;char ch=getchar();

	while (ch<'0'||ch>'9') {if (ch=='-') f=-1;ch=getchar();}

	while (ch>='0'&&ch<='9') {x=x*10+ch-48;ch=getchar();}

	return x*f;}

struct EDGE{int nxt,to,w;}e[2*N];

int n,m,Q,low[N],vis[N],fa[N][20],head[N],dpt[N],nd[MN],used[N],cnt=0,cnt2=0;

LL dist[N][MN],dist2[N];

inline void add(int u,int v,int w){e[++cnt].nxt=head[u];e[cnt].to=v;e[cnt].w=w;head[u]=cnt;}

void dfs(int u,int fath){

	used[u]=1;dpt[u]=dpt[fath]+1;fa[u][0]=fath;

	for (int i=1;i<=low[dpt[u]];++i){fa[u][i]=fa[fa[u][i-1]][i-1];}

	for (int i=head[u];i;i=e[i].nxt){

		int v=e[i].to,w=e[i].w;

		if (v==fath) continue;

		if (!used[v]) {dist2[v]=dist2[u]+w;dfs(v,u);}

			else{if (used[u]==1) nd[++cnt2]=u,used[u]++;

				 if (used[v]==1) nd[++cnt2]=v,used[v]++;}

	}

}

struct NODE{LL d;int u;bool operator< (const NODE &a)const{return d>a.d;}};

void dij(int stt,int now){

	for (int i=1;i<=n;++i) dist[i][now]=1e18+7;

	priority_queue<NODE> q;q.push((NODE){0,stt});dist[stt][now]=0;

	memset(vis,0,sizeof(vis));

	while (!q.empty()){

		NODE ntp=q.top();q.pop();

		int u=ntp.u;

		if (vis[u]) continue;

		vis[u]=1;

		for (int i=head[u];i;i=e[i].nxt){

			int v=e[i].to,w=e[i].w;

			if (dist[v][now]>dist[u][now]+w){

				dist[v][now]=dist[u][now]+w;

				q.push((NODE){dist[v][now],v});

			}

		}

	}

}

inline int lca(int x,int y){

	if (dpt[x]<dpt[y]) swap(x,y);

	while (dpt[x]>dpt[y]) {x=fa[x][low[dpt[x]-dpt[y]]];}

	if (x==y) return x;

	for (int i=low[dpt[x]];i>=0;--i)

		if (fa[x][i]!=fa[y][i]) x=fa[x][i],y=fa[y][i];

	return fa[x][0];}

int main(){

	int i,j;

	n=read();m=read();memset(head,0,sizeof(head));

	memset(used,0,sizeof(used));memset(dist2,0,sizeof(dist2));

	for (i=1;i<=m;++i){

		int u=read(),v=read(),w=read();add(u,v,w);add(v,u,w);}

	low[1]=0;for (i=2;i<=n;++i) low[i]=low[i>>1]+1;

	dpt[0]=0;dfs(1,0);

//	sort(nd+1,nd+1+cnt2);cnt2=unique(nd+1,nd+1+cnt2)-nd-1;

	for (i=1;i<=cnt2;++i) {int stt=nd[i];dij(stt,i);}

	Q=read();

	for (i=1;i<=Q;++i){

		int u=read(),v=read();

		int ast=lca(u,v);

		LL DIST=dist2[u]+dist2[v]-dist2[ast]*2;

		for (j=1;j<=cnt2;++j) DIST=Min(DIST,dist[u][j]+dist[v][j]);

		printf("%I64d\n",DIST);

	}

	return 0;

}