codefores544D

In some country there are exactly n cities and m bidirectional roads connecting the cities. Cities are numbered with integers from 1 to n. If cities a and b are connected by a road, then in an hour you can go along this road either from city a to city b, or from city b to city a. The road network is such that from any city you can get to any other one by moving along the roads.

You want to destroy the largest possible number of roads in the country so that the remaining roads would allow you to get from city s1 to city t1 in at most l1 hours and get from city s2 to city t2 in at most l2 hours.

Determine what maximum number of roads you need to destroy in order to meet the condition of your plan. If it is impossible to reach the desired result, print -1.

Input
The first line contains two integers n, m (1 ≤ n ≤ 3000, ) — the number of cities and roads in the country, respectively.

Next m lines contain the descriptions of the roads as pairs of integers ai, bi (1 ≤ ai, bi ≤ n, ai ≠ bi). It is guaranteed that the roads that are given in the description can transport you from any city to any other one. It is guaranteed that each pair of cities has at most one road between them.

The last two lines contains three integers each, s1, t1, l1 and s2, t2, l2, respectively (1 ≤ si, ti ≤ n, 0 ≤ li ≤ n).

Output
Print a single number — the answer to the problem. If the it is impossible to meet the conditions, print -1.

Example
Input
5 4
1 2
2 3
3 4
4 5
1 3 2
3 5 2
Output
0
Input
5 4
1 2
2 3
3 4
4 5
1 3 2
2 4 2
Output
1
Input
5 4
1 2
2 3
3 4
4 5
1 3 2
3 5 1
Output
-1

这题告诉我们如何用BFS当权值相同时n^2求任意两点间最短路，还有补集转化的力量

using namespace std;#include<bits/stdc++.h>#define N 3001#define forw(i,x) for(int i=fir[x];i;i=ne[i])int way[N][N];int n,m;int x,y;int s1,t1,l1,s2,t2,l2;int ne[N*2],fir[N],to[N*2],cnt=1;void add(int x,int y){    ne[++cnt]=fir[x];fir[x]=cnt;to[cnt]=y;}bool vis[N];queue<int>q;int res=2e9;int main(){    cin>>n>>m;    for(int i=1;i<=m;i++)    {        cin>>x>>y;add(x,y);add(y,x);    }    for(int i=1;i<=n;i++)    {        memset(vis,0,sizeof(vis));        vis[i]=1;way[i][i]=0;        while(!q.empty())q.pop();        q.push(i);        while(!q.empty())        {            int ind=q.front();q.pop();            forw(j,ind)            {                int v=to[j];                if(!vis[v])                {                    vis[v]=1;                    way[i][v]=way[i][ind]+1;                    q.push(v);                }            }        }    }    if(0)    for(int i=1;i<=n;i++)    {        for(int j=1;j<=n;j++)        {            cout<<way[i][j]<<" ";        }        puts("");    }    cin>>s1>>t1>>l1;    cin>>s2>>t2>>l2;    if(way[s1][t1]>l1||way[s2][t2]>l2)    {        puts("-1");return 0;    }    res=way[s1][t1]+way[s2][t2];//可能不选仅这一种可能，无相交     for(int i=1;i<=n;i++)    {        for(int j=1;j<=n;j++)        {            int it=way[i][j];            int jj=way[s1][i]+way[j][t1]+it;            int kk=way[s2][i]+way[j][t2]+it;            if(jj<=l1&&kk<=l2) res=min(res,jj+kk-it);            it=way[i][j];            jj=way[s1][i]+way[j][t1]+it;            kk=way[s2][j]+way[i][t2]+it;            if(jj<=l1&&kk<=l2) res=min(res,jj+kk-it);                   }    }    cout<<m-res<<endl;}