QwQ好久没更新博客了，颓废了好久啊，来补一点东西

题目大意

给定两个点对，求两对点间最短路的最长公共路径。

其中\(n,m\le 10^5\)

比较简单吧

就是跑四遍最短路，然后把最短路上的边拿出来，跑一遍拓扑排序加\(dp\)就OK

对于一条边\(u->v\)，满足\(dis[u]+w+disn[v]=dis[t]\)那么这条边就是最短路上的边，其中\(disn[x]\)表示\(t\)到\(x\)的最短路，\(dis[x]\)表示\(s\)到\(x\)的最短路

直接上代码

#include
    
     #include
     
      #include
      
       #include
       
        #include
        
         #include
         
          #define pa pair
          
            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
           
            
             ,greater
             
               > 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
              
                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<
               
                <<" "<
                
<<" "<
                 
                  <