F. The Shortest Statement
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-nleq 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
",DIST);
}
return 0;
}