QwQ好久沒更新部落格了，頹廢了好久啊，來補一點東西

題目大意

給定兩個點對，求兩對點間最短路的最長公共路徑。

其中\(n,m\le 10^5\)

比較簡單吧

就是跑四遍最短路，然後把最短路上的邊拿出來，跑一遍拓撲排序加\(dp\)就OK

對於一條邊\(u->v\)，滿足\(dis[u]+w+disn[v]=dis[t]\)那麼這條邊就是最短路上的邊，其中\(disn[x]\)表示\(t\)到\(x\)的最短路，\(dis[x]\)表示\(s\)到\(x\)的最短路

直接上程式碼

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#define pa pair<int,int> 
using namespace std;

inline int read()
{
  int x=0,f=1;char ch=getchar();
  while (!isdigit(ch)) {if (ch=='-') f=-1;ch=getchar();}
  while (isdigit(ch)) {x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
  return x*f;
}

const int maxn = 7010;
const int maxm = 2e6+1e2;

int point[maxn],nxt[maxm],to[maxm],val[maxm];
int dis[maxn];
int disn[maxn];
int dis1[maxn];
int disn1[maxn];
int vis[maxn];
int n,m,cnt;
int x[maxm],y[maxm],w[maxm];
int dp[maxn];
int s1,t1,s2,t2;
priority_queue<pa,vector<pa>,greater<pa> > q; 

void addedge(int x,int y,int w)
{
    nxt[++cnt]=point[x];
    to[cnt]=y;
    val[cnt]=w;
    point[x]=cnt;
}

void dijkstra(int s)
{
    memset(vis,0,sizeof(vis));
    memset(dis,127/3,sizeof(dis));
    dis[s]=0;
    q.push(make_pair(0,s));
    while (!q.empty())
    {
        int x = q.top().second;
        q.pop();
        if (vis[x]) continue;
        vis[x]=1;
        for (int i=point[x];i;i=nxt[i])
        {
            int p = to[i];
            if (dis[p]>dis[x]+val[i])
            {
                dis[p]=dis[x]+val[i];
                q.push(make_pair(dis[p],p));
             } 
        } 
    }
}

void dijkstra1(int s)
{
    memset(vis,0,sizeof(vis));
    memset(disn,127/3,sizeof(disn));
    disn[s]=0;
    q.push(make_pair(0,s));
    while (!q.empty())
    {
        int x = q.top().second;
        q.pop();
        if (vis[x]) continue;
        vis[x]=1;
        for (int i=point[x];i;i=nxt[i])
        {
            int p = to[i];
            if (disn[p]>disn[x]+val[i])
            {
                disn[p]=disn[x]+val[i];
                q.push(make_pair(disn[p],p));
             } 
        } 
    }
}

void dijkstra2(int s)
{
    memset(vis,0,sizeof(vis));
    memset(dis1,127/3,sizeof(dis1));
    dis1[s]=0;
    q.push(make_pair(0,s));
    while (!q.empty())
    {
        int x = q.top().second;
        q.pop();
        if (vis[x]) continue;
        vis[x]=1;
        for (int i=point[x];i;i=nxt[i])
        {
            int p = to[i];
            if (dis1[p]>dis1[x]+val[i])
            {
                dis1[p]=dis1[x]+val[i];
                q.push(make_pair(dis1[p],p));
             } 
        } 
    }
}

void dijkstra3(int s)
{
    memset(vis,0,sizeof(vis));
    memset(disn1,127/3,sizeof(disn1));
    disn1[s]=0;
    q.push(make_pair(0,s));
    while (!q.empty())
    {
        int x = q.top().second;
        q.pop();
        if (vis[x]) continue;
        vis[x]=1;
        for (int i=point[x];i;i=nxt[i])
        {
            int p = to[i];
            if (disn1[p]>disn1[x]+val[i])
            {
                disn1[p]=disn1[x]+val[i];
                q.push(make_pair(disn1[p],p));
             } 
        } 
    }
}

int tmp = 0;
int num[maxm];
int ymh[maxm];
int in[maxn];
queue<int> que;
int a[2100][2100];

int main()
{
  memset(a,-1,sizeof(a));
  n=read(),m=read();
  s1=read(),t1=read();
  s2=read(),t2=read();
  for (int i=1;i<=m;i++){
     int u,v,ww;
     u=read(),v=read(),ww=read();
     addedge(u,v,ww);
     addedge(v,u,ww);
     x[++tmp]=u;
     y[tmp]=v;
     w[tmp]=ww;
     x[++tmp]=v;
     y[tmp]=u;
     w[tmp]=ww;
  }
  dijkstra(s1);
  dijkstra1(t1);
  dijkstra2(s2);
  dijkstra3(t2);
  memset(point,0,sizeof(point));
  cnt=0;
  for (int i=1;i<=tmp;i++)
  {
     if (dis[x[i]]+w[i]+disn[y[i]]==dis[t1]) addedge(x[i],y[i],w[i]),num[i]=1,in[y[i]]++,a[x[i]][y[i]]++,a[y[i]][x[i]]++;
  }
  for (int i=1;i<=tmp;i++)
  {
     if (dis1[x[i]]+w[i]+disn1[y[i]]==dis1[t2]) {a[x[i]][y[i]]++,a[y[i]][x[i]]++;}
  }
  for (int i=1;i<=n;i++) if (!in[i]) que.push(i);
  while (!que.empty())
  {
     int x = que.front();
     que.pop();
     for (int i=point[x];i;i=nxt[i])
     {
        int p = to[i];
        //cout<<x<<" "<<p<<" "<<a[x][p]<<endl;
        dp[p]=max(dp[p],dp[x]+a[x][p]*val[i]);
        in[p]--;
        if (!in[p]) que.push(p);
       }
  }
  int ans=-1e9;
  for (int i=1;i<=n;i++) ans=max(ans,dp[i]);
  cout<<ans; 
  return 0;
}