Codeforces 1051F. The Shortest Statement

对于每一条边，如果是树边，直接加入，否则，考虑最终的最短路会不会经过此边，如果会经过，就依此边的任意一个节点为初始点，在包含所有边的图中跑最短路，dis[a]+dis[b]即可(a,b为所求路径的两个点)，dis[]为从该初始点出发到所有点最短距离的数组

在树边考虑最短路径时，需要用lca求最近公共祖先算法

#include<iostream>
#include<cstdio> 
#include<cmath>
#include<queue>
#include<vector>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<set>
#include<map>
#include<fstream>
#include<cstdlib>
#include<ctime>
#include<list>
#include<climits>
#include<bitset>
#include<random>
#include <ctime>
#include <cassert>
#include <complex>
#include <cstring>
#include <chrono>
using namespace std;
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("input.in", "r", stdin);freopen("output.in", "w", stdout);
#define left asfdasdasdfasdfsdfasfsdfasfdas1
#define tan asfdasdasdfasdfasfdfasfsdfasfdas
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
typedef long long ll;
typedef unsigned int un;
const int desll[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
const int mod=998244353;
const int maxn=1e5+7;
const int maxm=1e6+1;
const double eps=1e-8;
int n,k,m;
int ar[maxn*2];
struct node{
    int b,nex,c;
}no[maxn*2],noMid[maxn*2];
int sz,head[maxn],szMid,headMid[maxn];
int f[maxn];
int fin(int a)
{
    return a==f[a]?a:f[a]=fin(f[a]);
}
bool func(int a,int b){
    a=fin(a);
    b=fin(b);
    if(a==b)return 1;
    else {
        f[a]=b;
        return 0;
    }
}
void add(int a,int b,int c)
{
    no[sz].b=b;
    no[sz].c=c;
    no[sz].nex=head[a];
    head[a]=sz++;
}
void addMid(int a,int b,int c)
{
    noMid[szMid].b=b;
    noMid[szMid].c=c;
    noMid[szMid].nex=headMid[a];
    headMid[a]=szMid++;
}
int tan[maxn*2][40],in[maxn],dep[maxn],le;
ll dis[maxn];
void dfs(int u,int pre)
{
    for(int i=headMid[u];i!=-1;i=noMid[i].nex){
        int v=noMid[i].b;
        if(pre==v)continue;
        dep[v]=dep[u]+1;
        ar[le++]=u;
        dis[v]=dis[u]+noMid[i].c;
        dfs(v,u);
    }
    in[u]=le;
    ar[le++]=u;
}
void init()
{
    memset(tan,0,sizeof(tan));
    for(int i=0;i<le;i++){
        tan[i][0]=ar[i];
    }
    for(int j=1;j<40;j++){
        for(int i=0;i<le;i++){
            if(i+(1<<j)>le)break;
            int x=tan[i][j-1],y=tan[i+(1<<(j-1))][j-1];
            if(dep[x]<dep[y])tan[i][j]=x;
            else tan[i][j]=y;
        }
    }
}
int sameFather(int a,int b)
{
    a=in[a];b=in[b];
    if(a>b)swap(a,b);
    int res=floor(log(b-a+1)/log(2));
    int x=tan[a][res],y=tan[b-(1<<res)+1][res];
    if(dep[x]<dep[y])return x;
    else return y;
}
vector<int> ve;
ll disall[50][maxn];
bool vis[maxn];
priority_queue<pair<ll,int>>qu;
void dij(int i,int x)
{
    memset(disall[i],-1,sizeof(disall[i]));
    qu.push(make_pair(0,x));
    while(qu.size()){
        int u=qu.top().second;
        ll dd=-qu.top().first;qu.pop();
        if(disall[i][u]!=-1 && dd>disall[i][u])continue;
        disall[i][u]=dd;
        for(int j=head[u];j!=-1;j=no[j].nex){
            int v=no[j].b,c=no[j].c;
            if(disall[i][v]==-1 || dd+c<disall[i][v]){
                disall[i][v]=dd+c;
                qu.push(make_pair(-disall[i][v],v));
            }
        }
    }
}
int main()
{
    scanf("%d%d",&n,&m);
    memset(head,-1,sizeof(head));
    memset(headMid,-1,sizeof(headMid));
    for(int i=1;i<=n;i++)f[i]=i;
    sz=szMid=le=0;
    for(int i=0;i<m;i++){
        int a,b,c;scanf("%d%d%d",&a,&b,&c);
        add(a,b,c);
        add(b,a,c);
        if(!func(a,b)){
            addMid(a,b,c);
            addMid(b,a,c);
        }
        else ve.push_back(a);
    }
    dep[1]=1;
    dis[1]=0;
    dfs(1,-1);
    init();
    for(int i=0;i<ve.size();i++)dij(i,ve[i]);
    ll q,ans;
    scanf("%I64d",&q);
    while(q--){
        ans=1e15;
        int a,b;scanf("%d%d",&a,&b);
        int x=sameFather(a,b);
        ans = min(ans, dis[a]+dis[b]-dis[x]*2);
        for(int i=0;i<ve.size();i++){
            ans =min(ans, disall[i][a]+disall[i][b]);
        }
        printf("%I64d
",ans);
    }
    return 0;
}