codeforces 1051 F

 

多源点最短路。

 

但是有限制，m - n <= 20，边数 - 点数 <= 20， 所以这个图非常的稀疏。

 

任意提取出一个生成树出来，LCA处理任意两点最短路。

 

然后再去遍历那些多余出来的点（非树边的两个端点），看他们能不能更新答案（相当于松弛）。这里跑40个最短路预处理出来（最多40个点，但是必须在原图上跑才行）。

 

 

 

#include <bits/stdc++.h>

using namespace std;

#define time _____time
const long long inf = 1E18;
struct edge {
    long long u,v,val,next;
};

edge Edge[201000];
edge xdge[201000];

int e = 0;
int head[100005];

void addedge(int u,int v,int val) {
    Edge[e].u = u;
    Edge[e].v = v;
    Edge[e].val = val;
    Edge[e].next = head[u];
    head[u] = e++;

    Edge[e].u = v;
    Edge[e].v = u;
    Edge[e].val = val;
    Edge[e].next = head[v];
    head[v] = e++;

}

int ec = 0;
int head2[100005];

void ADD(int u,int v,int val) {
    xdge[ec].u = u;
    xdge[ec].v = v;
    xdge[ec].val = val;
    xdge[ec].next = head2[u];
    head2[u] = ec++;

    xdge[ec].u = v;
    xdge[ec].v = u;
    xdge[ec].val = val;
    xdge[ec].next = head2[v];
    head2[v] = ec++;
}

int in[100005];
int out[100005];
int time = 0;
long long depth[100005];
long long ST[200005*2][20];
void dfs(int x,int pre)
{
    in[x]=++time;
    ST[time][0]=x;
    for (int i = head2[x];i != -1;i = xdge[i].next)
        if (xdge[i].v != pre){
            depth[xdge[i].v] = depth[x] + xdge[i].val;
            dfs(xdge[i].v,x);
            ST[++time][0]=x;
        }
    out[x]=time;
}

int n,m;

void Get_ST(int n){
    for (int i=1;i<=n;i++)
        for (int j=1;j<20;j++){
            ST[i][j]=ST[i][j-1];
            int v=i-(1<<(j-1));
            if (v>0&&depth[ST[v][j-1]]<depth[ST[i][j]])
                ST[i][j]=ST[v][j-1];
        }
}

int RMQ(int L,int R){
    if(L > R) {
        swap(L,R);
    }
    int val=floor(log(R-L+1)/log(2));
    int x=ST[L+(1<<val)-1][val],y=ST[R][val];
    if (depth[x]<depth[y])
        return x;
    else
        return y;
}

int fa[100005];
int ffind(int x) {
    if(fa[x] == x) return x;
    else return fa[x] = ffind(fa[x]);
}
void unit(int x,int y) {
    int fx,fy;
    fx = ffind(x);
    fy = ffind(y);
    if(fx != fy) {
        fa[fx] = fy;
    }
}

struct bian {
    int l,r,val;
};

vector<bian> duo;

long long mmap[45][45];
map<int,int> mp;
vector<int> dian;

long long dis[45][100005];
int vis[100005];

typedef pair<long long,int> pli;

void dij(int id,int s)
{
    for(int i=0;i<=n+2;i++){
        vis[i]=0;
        dis[id][i]=inf;
    }
    dis[id][s]=0;
    priority_queue<pli,vector<pli>,greater<pli> >q;
    q.push(pli(0,s));

    while(!q.empty())
    {
        pli tmp=q.top();
        q.pop();
        int u=tmp.second;
        if(tmp.first>dis[id][u]) continue;
        vis[u]=1;

        for(int i=head[u];i!=-1;i = Edge[i].next){
            int v=Edge[i].v;
            long long w=Edge[i].val;
            if(vis[v]) continue;
            if(dis[id][v]>dis[id][u]+w)
            {
                dis[id][v]=dis[id][u]+w;
                q.push(pli(dis[id][v],v));
            }
        }
    }

}


int main() {

    scanf("%d %d",&n,&m);
    int u,v;
    for(int i = 0; i <= n; i++) {
        fa[i] = i;
        head[i] = -1;
        head2[i] = -1;
    }
    bian xx;
    int val;
    for(int i = 1; i <= m; i++) {
        scanf("%d %d %d",&u,&v,&val);
        addedge(u,v,val);
        if(ffind(u) != ffind(v)) {
            unit(u,v);
            ADD(u,v,val);
        }
        else {
            dian.push_back(u);
            dian.push_back(v);
        }
    }


    sort(dian.begin(),dian.end());
    dian.erase(unique(dian.begin(),dian.end()),dian.end());

    for(int i=0;i<dian.size();i++){
        //cout<<dian[i]<<endl;
        dij(i + 1,dian[i]);
    }

    dfs(1,0);
    depth[0]= inf;

    Get_ST(time);


    int q;
    scanf("%d",&q);
    while (q--){

        int x,y;
        scanf("%d %d",&x,&y);
        int LCA=RMQ(in[x],in[y]);
        long long ans = depth[x]+depth[y]-depth[LCA]*2;
        for(int i = 0; i < dian.size(); i++) {
            ans = min(ans,dis[i + 1][x] + dis[i + 1][y]);
        }

        printf("%lld\n",ans);



    }







}