http://www.elijahqi.win/archives/1612
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
这题要求是 我们s1~t1 的最短路<=l1
要求我s2~t2的最短路<=l2
求除了这些边以外 最多可以删掉多少边
那我可以宽搜抛出多源最短路
我每次去暴力枚举他们两条路径的重边 然后求出最多可以删去多少
注意枚举的时候一共是四种情况需要枚举 即s1 i j t1 s2 i j t2
s1 j i t1 s2 j i t2
s1 i j t1 s2 j i t2
s1 j i t1 s2 i j t2
#include<cstdio>
#define N 3300
#include<queue>
#include<cstring>
#define inf 0x3f3f3f3f
using namespace std;
inline char gc(){
static char now[1<<16],*S,*T;
if (T==S){T=(S=now)+fread(now,1,1<<16,stdin);if (T==S) return EOF;}
return *S++;
}
inline int read(){
int x=0;char ch=gc();
while (ch<'0'||ch>'9') ch=gc();
while (ch<='9'&&ch>='0'){x=x*10+ch-'0';ch=gc();}
return x;
}
struct node{
int y,next;
}data[N<<1];
int h[N],dis[N][N],s1,s2,t1,t2,l1,l2,n,m,num;bool flag[N];
inline void bfs(int fa,int u){
queue<int> q;q.push(u);memset(flag,0,sizeof(flag));flag[fa]=1;
while(!q.empty()){
int x=q.front();q.pop();
for (int i=h[x];i;i=data[i].next){
int y=data[i].y;if (!flag[y]){
dis[fa][y]=dis[fa][x]+1;q.push(y),flag[y]=1;
}
}
}
}
int main(){
freopen("cf.in","r",stdin);
n=read();m=read();
for (int i=1;i<=m;++i){
int x=read(),y=read();data[++num].y=y;data[num].next=h[x];h[x]=num;
data[++num].y=x;data[num].next=h[y];h[y]=num;
}
for (int i=1;i<=n;++i) bfs(i,i);
s1=read();t1=read();l1=read();
s2=read();t2=read();l2=read();int ans=m-(dis[s1][t1]+dis[s2][t2]);
if (dis[s1][t1]>l1||dis[s2][t2]>l2) {printf("-1");return 0;}
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){
ans=max(ans,m-(dis[i][j]+dis[s1][i]+dis[j][t1]+dis[s2][i]+dis[j][t2]));
}
if (dis[s1][i]+dis[i][j]+dis[j][t1]<=l1&&dis[s2][j]+dis[j][i]+dis[i][t2]<=l2){
ans=max(ans,m-(dis[i][j]+dis[s1][i]+dis[j][t1]+dis[s2][j]+dis[i][t2]));
}
}
}
printf("%d",ans);
return 0;
}