cf1051F. The Shortest Statement（最短路/dfs树）

You are given a weighed undirected connected graph, consisting of $\mathrm{nn vertices\; and mm edges.}$

You should answer $\mathrm{qq queries,\; the ii-th\; query\; is\; to\; find\; the\; shortest\; distance\; between\; vertices uiui and vivi.}$

Input

The first line contains two integers $\mathrm{n and m (1\le n,m\le 105,m-n\le 20) —\; the\; number\; of\; vertices\; and\; edges\; in\; the\; graph.}$

Next $\mathrm{m lines\; contain\; the\; edges:\; the i-th\; edge\; is\; a\; triple\; of\; integers vi,ui,di (1\le ui,vi\le n,1\le di\le 109,ui\ne vi).\; This\; triple\; means\; that\; there\; is\; an\; edge\; between\; vertices ui and vi of\; weight di.\; It\; is\; guaranteed\; that\; graph\; contains\; no\; self-loops\; and\; multiple\; edges.}$

The next line contains a single integer $\mathrm{q (1\le q\le 105)—\; the\; number\; of\; queries.}$

Each of the next $\mathrm{qq lines\; contains\; two\; integers uiui and vi (1\le ui,vi\le n)vi (1\le ui,vi\le n) —\; descriptions\; of\; the\; queries.}$

Pay attention to the restriction $\mathrm{m-n \le 20}$

Output

Print $\mathrm{q lines.}$

The $\mathrm{i-th\; line\; should\; contain\; the\; answer\; to\; the i-th\; query\; —\; the\; shortest\; distance\; between\; vertices ui and vi.}$

$\mathrm{解题思路：}$

题目非常良心地强调了m-n<=20

建一颗dfs树，对于未加入树的边两端跑Dij

询问只要枚举未入树边更新即可。

代码：

#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
typedef long long lnt;
struct pnt{
    int hd;
    int dp;
    int no;
    int fa[22];
    lnt dis;
    lnt tds;
    bool vis;
    bool cop;
    bool friend operator < (pnt x,pnt y)
    {
        return x.dis>y.dis;
    }
}p[1000000];
struct ent{
    int twd;
    int lst;
    lnt vls;
}e[1000000];
std::priority_queue<pnt>Q;
int n,m;
int q;
int cnt;
int sct;
lnt dist[50][200000];
int sta[50];
void ade(int f,int t,lnt v)
{
    cnt++;
    e[cnt].twd=t;
    e[cnt].vls=v;
    e[cnt].lst=p[f].hd;
    p[f].hd=cnt;
    return ;
}
void Dij(int x)
{
    if(p[x].cop)
        return ;
    p[x].cop=true;
    sta[++sct]=x;
    for(int i=1;i<=n;i++)
    {
        p[i].dis=0x3f3f3f3f3f3f3f3fll;
        p[i].no=i;
        p[i].vis=false;
    }
    p[x].dis=0;
    while(!Q.empty())
        Q.pop();
    Q.push(p[x]);
    while(!Q.empty())
    {
        x=Q.top().no;
        Q.pop();
        if(p[x].vis)
            continue;
        p[x].vis=true;
        for(int i=p[x].hd;i;i=e[i].lst)
        {
            int to=e[i].twd;
            if(p[to].dis>p[x].dis+e[i].vls)
            {
                p[to].dis=p[x].dis+e[i].vls;
                Q.push(p[to]);
            }
        }
    }
    for(int i=1;i<=n;i++)
        dist[sct][i]=p[i].dis;
    return ;
}
void dfs(int x,int f)
{
    p[x].fa[0]=f;
    p[x].dp=p[f].dp+1;
    for(int i=1;i<=20;i++)
        p[x].fa[i]=p[p[x].fa[i-1]].fa[i-1];
    for(int i=p[x].hd;i;i=e[i].lst)
    {
        int to=e[i].twd;
        if(to==f)
            continue;
        if(p[to].tds)
        {
            Dij(x);
            Dij(to);
        }else{
            p[to].tds=p[x].tds+e[i].vls;
            dfs(to,x);
        }
    }
}
int Lca(int x,int y)
{
    if(p[x].dp<p[y].dp)
        std::swap(x,y);
    for(int i=20;i>=0;i--)
        if(p[p[x].fa[i]].dp>=p[y].dp)
            x=p[x].fa[i];
    if(x==y)
        return x;
    for(int i=20;i>=0;i--)
        if(p[x].fa[i]!=p[y].fa[i])
        {
            x=p[x].fa[i];
            y=p[y].fa[i];
        }
    return p[x].fa[0];
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++)
    {
        int a,b,c;
        scanf("%d%d%d",&a,&b,&c);
        ade(a,b,c);
        ade(b,a,c);
    }
    p[1].tds=1;
    dfs(1,1);
    scanf("%d",&q);
    while(q--)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        int f=Lca(x,y);
        lnt ans=p[x].tds+p[y].tds-2*p[f].tds;
        for(int i=1;i<=sct;i++)
            ans=std::min(ans,dist[i][x]+dist[i][y]);
        printf("%I64d\n",ans);
    }
    return 0;
}