Description

N has an unrooted tree nodes, wherein the given point m of <x, y>. I asked how many simple path tree <u, v> does not exist on the path to meet any given pair of points of <x, y>.
Here we consider the path <u, v> and <v, u> is the same. For the given point and the title of <x, y> satisfies x! = Y, the path <u, v> you want to count satisfies u! = V (i.e., not a single point answer).

Input

The first line of two positive integers n, m.
Next row n-1 in each row two positive integers u, v describe a tree edge.
The next two lines each line m a positive integer x, y description is given of a pair of points.
(Note that here we do not guarantee the same point <x, y> will not be repeated)

Output

Line an integer representing the number of paths of the tree simply to meet the requirements.

Sample Input

Sample Output

Data Constraint

Hint

Satisfies the path condition <1,2>, <1,3>, <1,4>, <1,5>, <1,6>, <1,7>, <2,4>, <2, 5>, <3,6>, <3,7>, <4,5>.

analysis

We consider it DFS processing sequence, then obviously, when the two points can not elect not to ancestral relationship, which does not have the legal number of paths in the subtree size product

We call this a two-dimensional plane of the abstract, is the greatest sequence DFS DFS sequence and a point in the subtree as wide, as another similar long rectangular area

We have put all the points on the plane, the total area is the number of illegal schemes, can be treated with the scan line segment tree +

Ancestral relationships is similar, except ancestor son comprising subtrees can not be selected, it is [1, l [son] -1] [r [son] + 1, n] of the two bounding rectangle

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int N=1e5+10;
struct Line {
    int x,l,r,c;
}h[4*N];
int lcnt;
struct Graph {
    int v,nx;
}g[2*N];
int cnt,list[N],l[N],r[N],tme,f[N][20],dep[N],t[4*N],v[4*N];
int n,m;
long long ans;

void Add(int u,int v) {
    g[++cnt]=(Graph){v,list[u]};list[u]=cnt;
    g[++cnt]=(Graph){u,list[v]};list[v]=cnt;
}

void Addline(int x0,int x1,int y0,int y1) {
    h[++lcnt]=(Line){x0,y0,y1,1};h[++lcnt]=(Line){x1+1,y0,y1,-1};
}

void DFS(int u,int fa) {
    f[u][0]=fa;l[u]=++tme;dep[u]=dep[fa]+1;
    for (int i=list[u];i;i=g[i].nx)
        if (g[i].v!=fa) DFS(g[i].v,u);
    r[u]=tme;
}

int LCA(int a,int b) {
    if (dep[a]<dep[b]) swap(a,b);
    for (int i=19;i>=0;i--)
        if (dep[f[a][i]]>=dep[b]) a=f[a][i];
    if (a==b) return a;
    for (int i=19;i>=0;i--)
        if (f[a][i]!=f[b][i]) a=f[a][i],b=f[b][i];
    return f[a][0];
}

bool CMP(Line a,Line b) {
    return a.x<b.x;
}

void Change(int x,int l,int r,int ll,int rr,int c) {
    if (r<l||rr<l||r<ll) return;
    if (ll<=l&&r<=rr) {
        t[x]+=c;
        if (t[x]) v[x]=r-l+1;
        else if (l!=r) v[x]=v[x<<1]+v[(x<<1)+1];
        else v[x]=0;
        return;
    }
    int mid=l+r>>1;
    if (ll<=mid) Change(x<<1,l,mid,ll,rr,c);
    if (mid<rr) Change((x<<1)+1,mid+1,r,ll,rr,c);
    v[x]=t[x]?r-l+1:v[x<<1]+v[(x<<1)+1];
}

int main() {
    freopen("tree.in","r",stdin);
    freopen("tree.out","w",stdout);
    scanf("%d%d",&n,&m);
    for (int i=1,u,v;i<n;i++) scanf("%d%d",&u,&v),Add(u,v);
    DFS(1,0);
    for (int i=1;i<20;i++)
        for (int j=1;j<=n;j++) f[j][i]=f[f[j][i-1]][i-1];
    for (int i=1,x,y;i<=m;i++) {
        scanf("%d%d",&x,&y);
        if (l[x]>l[y]) swap(x,y);
        int lca=LCA(x,y);
        if (lca==x) {
            int s=y;
            for (int i=19;i>=0;i--)
                if (dep[f[s][i]]>dep[x]) s=f[s][i];
            if (l[s]-1>0) Addline(1,l[s]-1,l[y],r[y]);
            if (r[s]+1<=n) Addline(l[y],r[y],r[s]+1,n);
        }
        else Addline(l[x],r[x],l[y],r[y]);
    }
    sort(h+1,h+lcnt+1,CMP);
    for (int i=1,j=1;i<=n;i++) {
        while (h[j].x==i&&j<=lcnt) Change(1,1,n,h[j].l,h[j].r,h[j].c),j++;
        ans+=v[1];
    }
    printf("%lld\n",1ll*n*(n-1ll)/2ll-ans);
}

View Code

[Scanning line] [segment tree] JZOJ 6276 tree