Codeforces 543 B. World Tour

http://codeforces.com/problemset/problem/543/B

题意：

给定一张边权均为1的无向图。

问至多可以删除多少边，使得s1到t1的最短路不超过l1，s2到t2的最短路不超过l2。

 

转化成至少保留多少条边

若两条路径没有没有交集，就是dis[a1][b1]+dis[a2][b2]

如果两条路径有交集，交集一定是连续的一段，枚举交集的起点和终点即可

注意s1向t1方向可能对应着s2向t2方向，也可能对应着t2向s2方向

所以要交换s1 t1 求两遍

#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>

using namespace std;

#define N 3001

int dis[N][N];

int s1,t1,s2,t2,l1,l2;

int n;

int tot;
int front[N],to[N*N],nxt[N*N];

int mi;

void read(int &x)
{
    x=0; char c=getchar();
    while(!isdigit(c)) c=getchar();
    while(isdigit(c)) { x=x*10+c-'0'; c=getchar(); }
}

void add(int u,int v)
{
    to[++tot]=v; nxt[tot]=front[u]; front[u]=tot;
    to[++tot]=u; nxt[tot]=front[v]; front[v]=tot;
}

void bfs(int s)
{
    int now,t;
    static queue<int>q;
    q.push(s);
    while(!q.empty())
    {
        now=q.front();
        q.pop();
        for(int i=front[now];i;i=nxt[i])
        {
            t=to[i];
            if(dis[s][t]==-1) 
            {
                dis[s][t]=dis[s][now]+1;
                q.push(t);
            }
        }
    }
}

void solve()
{
    for(int i=1;i<=n;++i)
        for(int 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) continue;
            mi=min(mi,dis[s1][i]+dis[j][t1]+dis[s2][i]+dis[j][t2]+dis[i][j]);
            
        }
}

int main()
{
    int m;
    read(n); read(m);
    int u,v;
    for(int i=1;i<=m;++i)
    {
        read(u); read(v);
        add(u,v);
    }
    memset(dis,-1,sizeof(dis));
    for(int i=1;i<=n;++i) dis[i][i]=0;
    for(int i=1;i<=n;++i) bfs(i);
    read(s1); read(t1); read(l1);
    read(s2); read(t2); read(l2);
    if(dis[s1][t1]>l1 || dis[s2][t2]>l2)
    {
        printf("-1");
        return 0;
    }
    mi=dis[s1][t1]+dis[s2][t2];
    solve();
    swap(s1,t1);
    solve();
    printf("%d",m-mi);
}

B. Destroying Roads

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

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