codeforces round# 302(div 2 D)(思路)

本题目要让删除尽量多的边，使得图上存在一条从s1 -> t1 不超过 l1 的路径，和一条从s2 -> t2 不超过 l2 的路径。因为点数不超过3000，边数也不超过3000，且边权都为1

那么直接bfs求出任意两点的最短路径。

然后直接枚举重叠部分的两个点，重叠的起点和重叠终点然后暴力就可以了。

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <string>
#include <list>
#include <cstdlib>
#include <queue>
#include <stack>
#include <cmath>
#include <bitset>
#include <cassert>
#define ALL(a) a.begin(), a.end()
#define clr(a, x) memset(a, x, sizeof a)
#define fst first
#define snd second
#define pb push_back
#define lowbit(x) (x&(-x))
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define rep1(i,x,y) for(int i=x;i<=y;i++)
#define rep(i,n) for(int i=0;i<(int)n;i++)
using namespace std;
const double eps = 1e-10;
typedef long long LL;
typedef long long ll;
typedef pair<int, int> pii;
const int oo =0x3f3f3f3f;

const int N = 3030;
vector<int> G
;
int d

,n,m;
void bfs(int s){
   queue<int> Q;
   Q.push(s);
   clr(d[s],oo); d[s][s] = 0;
   while(!Q.empty()){
      int u = Q.front(); Q.pop();
      rep(i,G[u].size()) if(d[s][G[u][i]]>d[s][u]+1){
          d[s][G[u][i]] = d[s][u]+1;
          Q.push(G[u][i]);
      }
   }
}
int s1,t1,l1,s2,t2,l2;
int main()
{
   scanf("%d %d",&n,&m);
   rep1(i,1,m){
      int x,y; scanf("%d %d",&x,&y);
      G[x].push_back(y);
      G[y].push_back(x);
   }
   scanf("%d %d %d",&s1,&t1,&l1);
   scanf("%d %d %d",&s2,&t2,&l2);
   rep1(i,1,n) bfs(i);
   if(d[s1][t1] > l1 || d[s2][t2] > l2){
       printf("-1\n"); return 0;
   }
   int ans = d[s1][t1] + d[s2][t2];
   rep1(i,1,n) rep1(j,1,n){
       if(d[i][j]==oo) continue;
       if(d[s1][i]+d[i][j]+d[j][t1]<=l1 && d[s2][i]+d[i][j]+d[j][t2]<=l2)
           ans = min(ans,d[s1][i]+d[i][j]+d[j][t1]+d[s2][i]+d[j][t2]);
       if(d[s1][i]+d[i][j]+d[j][t1]<=l1 && d[t2][i]+d[i][j]+d[j][s2]<=l2)
           ans = min(ans,d[s1][i]+d[i][j]+d[j][t1]+d[t2][i]+d[j][s2]);
   }
   printf("%d\n",m - ans);
   return 0;
}