Mr. Kitayuta's Colorful Graph CodeForces - 506D（均摊复杂度）

Mr. Kitayuta has just bought an undirected graph with n vertices and m edges. The vertices of the graph are numbered from 1 to n. Each edge, namely edge i, has a color c_i, connecting vertex a_i and b_i.

Mr. Kitayuta wants you to process the following q queries.

In the i-th query, he gives you two integers - u_i and v_i.

Find the number of the colors that satisfy the following condition: the edges of that color connect vertex u_i and vertex v_i directly or indirectly.

Input

The first line of the input contains space-separated two integers - n and m(2 ≤ n ≤ 10⁵, 1 ≤ m ≤ 10⁵), denoting the number of the vertices and the number of the edges, respectively.

The next m lines contain space-separated three integers - a_i, b_i(1 ≤ a_i < b_i ≤ n) and c_i(1 ≤ c_i ≤ m). Note that there can be multiple edges between two vertices. However, there are no multiple edges of the same color between two vertices, that is, if i ≠ j, (a_i, b_i, c_i) ≠ (a_j, b_j, c_j).

The next line contains a integer- q(1 ≤ q ≤ 10⁵), denoting the number of the queries.

Then follows q lines, containing space-separated two integers - u_i and v_i(1 ≤ u_i, v_i ≤ n). It is guaranteed that u_i ≠ v_i.

Output

For each query, print the answer in a separate line.

Examples

Input

4 5
1 2 1
1 2 2
2 3 1
2 3 3
2 4 3
3
1 2
3 4
1 4

Output

2
1
0

Input

5 7
1 5 1
2 5 1
3 5 1
4 5 1
1 2 2
2 3 2
3 4 2
5
1 5
5 1
2 5
1 5
1 4

Output

1
1
1
1
2

Note

Let's consider the first sample.

The figure above shows the first sample.

• Vertex 1 and vertex 2 are connected by color 1 and 2.
• Vertex 3 and vertex 4 are connected by color 3.
• Vertex 1 and vertex 4 are not connected by any single color.

 

 

题意：给定N点M边无向图，每边有自己的颜色，Q次询问，每次给出(u,v)，询问多少种颜色，使得u和v连通。

思路：排序，同一种颜色同时处理，然后离线回答每个询问，但是有可以一个点存在M种颜色里，而且存在Q次询问里，所以最坏情况是M*Q*log。log是并查集的复杂度。所以要加均摊。 如果一种颜色的点比较多，就上面那么回答； 否则，我们就暴力记录点对。

总的复杂度是Q*sqrt(N)*log(N); 4s可以过了； 实际上只跑了500ms，还可以。

#include<bits/stdc++.h>
#define pii pair<int,int>
#define mp make_pair
#define F first
#define S second
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn=200010;
int Laxt[maxn],Next[maxn],To[maxn],id[maxn],ans[maxn],cnt;
int a[maxn],b[maxn],fa[maxn],times[maxn],T,q[maxn],tot,N;
map<pii,int>Mp;
map<pii,int>fcy;
struct in{
    int u,v,col;
    bool friend operator <(in w,in v){ return w.col<v.col; }
}s[maxn];
void add(int u,int v,int o)
{
    Next[++cnt]=Laxt[u]; Laxt[u]=cnt; To[cnt]=v; id[cnt]=o;
}
int find(int x){
    if(x==fa[x]) return x;
    return fa[x]=find(fa[x]);
}
void merge(int x,int y)
{
    int fx=find(x),fy=find(y);
    fa[fx]=fy;
}
int main()
{
    int M,Q,u,v;
    scanf("%d%d",&N,&M);
    rep(i,1,M) scanf("%d%d%d",&s[i].u,&s[i].v,&s[i].col);
    sort(s+1,s+M+1);
    scanf("%d",&Q);
    rep(i,1,Q){
        scanf("%d%d",&a[i],&b[i]); if(a[i]>b[i]) swap(a[i],b[i]);
        add(a[i],b[i],i);
        fcy[mp(a[i],b[i])]=1;
    }
    rep(i,1,M){
        int j=i; T++; merge(s[i].u,s[i].v);
        while(j+1<=M&&s[j+1].col==s[i].col) j++;
        tot=0;
        rep(k,i,j) q[++tot]=s[k].u,q[++tot]=s[k].v;
        sort(q+1,q+tot+1); tot=unique(q+1,q+tot+1)-(q+1);
        rep(k,1,tot) fa[q[k]]=q[k],times[q[k]]=T;
        rep(k,i,j) merge(s[k].u,s[k].v);
        if(tot>sqrt(N)) rep(k,1,tot) {
            for(int w=Laxt[q[k]];w;w=Next[w]){
                if(times[To[w]]==T&&find(q[k])==find(To[w])) ans[id[w]]++;
            }
        }
        else {
            rep(k,1,tot)
             rep(p,k+1,tot){
                  if(find(q[k])==find(q[p])&&fcy.find(mp(q[k],q[p]))!=fcy.end()) Mp[mp(q[k],q[p])]++;
            }
        }
        i=j;
    }
    rep(i,1,Q) printf("%d\n",ans[i]+Mp[mp(a[i],b[i])]);
    return 0;
}

可撤销并查集版本，530ms

#include<bits/stdc++.h>
#define pii pair<int,int>
#define mp make_pair
#define F first
#define S second
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn=1000010;
int Laxt[maxn],Next[maxn],To[maxn],id[maxn],ans[maxn],cnt;
int a[maxn],b[maxn],fa[maxn],times[maxn],T,q[maxn],tot,N;
map<pii,int>Mp,fcy;
struct in{
    int u,v,col;
    bool friend operator <(in w,in v){ return w.col<v.col; }
}s[maxn];
void add(int u,int v,int o)
{
    Next[++cnt]=Laxt[u]; Laxt[u]=cnt; To[cnt]=v; id[cnt]=o;
}
int find(int x){
    if(times[x]!=T) times[x]=T, fa[x]=x;
    if(x==fa[x]) return x;
    return fa[x]=find(fa[x]);
}
void merge(int x,int y)
{
    int fx=find(x),fy=find(y);
    fa[fx]=fy;
}
int main()
{
    int M,Q,u,v;
    scanf("%d%d",&N,&M);
    rep(i,1,M) scanf("%d%d%d",&s[i].u,&s[i].v,&s[i].col);
    sort(s+1,s+M+1);
    scanf("%d",&Q);
    rep(i,1,Q){
        scanf("%d%d",&a[i],&b[i]); if(a[i]>b[i]) swap(a[i],b[i]);
        add(a[i],b[i],i);
        fcy[mp(a[i],b[i])]=1;
    }
    rep(i,1,M){
        int j=i; T++; merge(s[i].u,s[i].v);
        while(j+1<=M&&s[j+1].col==s[i].col) j++;
        tot=0;
        rep(k,i,j) q[++tot]=s[k].u,q[++tot]=s[k].v,merge(s[k].u,s[k].v);;
        sort(q+1,q+tot+1); tot=unique(q+1,q+tot+1)-(q+1);
        if(tot>sqrt(N)) rep(k,1,tot) {
            for(int w=Laxt[q[k]];w;w=Next[w]){
                if(times[To[w]]==T&&find(q[k])==find(To[w])) ans[id[w]]++;
            }
        }
        else {
            rep(k,1,tot)
             rep(p,k+1,tot){
                  if(find(q[k])==find(q[p])&&fcy.find(mp(q[k],q[p]))!=fcy.end()) Mp[mp(q[k],q[p])]++;
            }
        }
        i=j;
    }
    rep(i,1,Q) printf("%d\n",ans[i]+Mp[mp(a[i],b[i])]);
    return 0;
}