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看题解才懂..

首先可以肯定的是 除了两条最短路上的边 其余边都可以去掉

而这两条最短路上的边 是否可以合并某些道路 使距离仍在l1与l2范围内(合并后未必是最短路)

求任意两点最短路 暴力枚举任意两点 看合并后是否满足l1与l2的范围

注意对每一对枚举的点i与j 有四种情况 s1可以连到i或j s2也可以连到i或j

#include <bits/stdc++.h>
using namespace std;

struct node1
{
  int v;
  int w;
  int next;
};

struct node2
{
  bool friend operator < (node2 n1,node2 n2)
  {
    return n1.w>n2.w;
  }
  int v;
  int w;
};

node1 edge[1000000];
priority_queue <node2> que;
int dis[3010][3010];
int first[3010],book[3010];
int n,m,num;
int s1,t1,l1,s2,t2,l2;

void addedge(int u,int v,int w)
{
  edge[num].v=v;
  edge[num].w=w;
  edge[num].next=first[u];
  first[u]=num++;
  return;
}

void dijkstra()
{
  node2 cur,tem;
  int i,j,u,v,w;
  memset(dis,0x3f,sizeof(dis));
  for(i=1;i<=n;i++)
  {
    while(!que.empty()) que.pop();
    memset(book,0,sizeof(book));
    tem.v=i,tem.w=0;
    que.push(tem);
    dis[i][i]=0;
    while(!que.empty())
    {
      cur=que.top();
      que.pop();
      u=cur.v;
      if(book[u]) continue;
      book[u]=1;
      for(j=first[u];j!=-1;j=edge[j].next)
      {
        v=edge[j].v,w=edge[j].w;
        if(!book[v]&&dis[i][v]>dis[i][u]+w)
        {
          dis[i][v]=dis[i][u]+w;
          tem.v=v,tem.w=dis[i][v];
          que.push(tem);
        }
      }
    }
  }
  return;
}

int main()
{
  int i,j,u,v,ans;
  while(scanf("%d%d",&n,&m)!=EOF)
  {
    memset(first,-1,sizeof(first));
    num=0;
    for(i=1;i<=m;i++)
    {
      scanf("%d%d",&u,&v);
      addedge(u,v,1);
      addedge(v,u,1);
    }
    scanf("%d%d%d",&s1,&t1,&l1);
    scanf("%d%d%d",&s2,&t2,&l2);
    dijkstra();
    if(!(dis[s1][t1]<=l1&&dis[s2][t2]<=l2))
    {
      printf("-1\n");
      continue;
    }
    ans=dis[s1][t1]+dis[s2][t2];
    for(i=1;i<=n;i++)
    {
      for(j=1;j<=n;j++)
      {
        if(dis[s1][i]+dis[i][j]+dis[j][t1]<=l1&&dis[s2][i]+dis[i][j]+dis[j][t2]<=l2)
        {
          ans=min(ans,dis[s1][i]+dis[j][t1]+dis[s2][i]+dis[j][t2]+dis[i][j]);
        }
        if(dis[s1][i]+dis[i][j]+dis[j][t1]<=l1&&dis[t2][i]+dis[i][j]+dis[j][s2]<=l2)
        {
          ans=min(ans,dis[s1][i]+dis[j][t1]+dis[t2][i]+dis[j][s2]+dis[i][j]);
        }
        if(dis[t1][i]+dis[i][j]+dis[j][s1]<=l1&&dis[s2][i]+dis[i][j]+dis[j][t2]<=l2)
        {
          ans=min(ans,dis[t1][i]+dis[j][s1]+dis[s2][i]+dis[j][t2]+dis[i][j]);
        }
        if(dis[t1][i]+dis[i][j]+dis[j][s1]<=l1&&dis[t2][i]+dis[i][j]+dis[j][s2]<=l2)
        {
          ans=min(ans,dis[t1][i]+dis[j][s1]+dis[t2][i]+dis[j][s2]+dis[i][j]);
        }
      }
    }
    printf("%d\n",m-ans);
  }
  return 0;
}