Route planning

Fates finally blessing me! ! !

I played two days cost flow, not WA stuck rotation is inexplicable TLE, this question that I am actually at arm's length over AC! ! !

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 Into the title.

Considering: Always be up to each other, knowing that it will never Unicom.

Truth: plus side is always better to do than delete the edge, so it is natural to think off.

Practice: 1. to find a spanning tree in all remaining sides.

　　    2. The remaining side is not coupled in the spanning tree.

　　　3. Enter the opposite of statistical answers or edging.

　　    4. flashback output answer, end Sahua.

It is so easy. . . . . .

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e6;
int n,m;
int beg[maxn],nex[maxn],to[maxn],e;
inline void add(int x,int y){
    e++;nex[e]=beg[x];
    beg[x]=e;to[e]=y;
}
struct node{
    int x,y,flag;
}a[maxn],q[maxn];
an int FA [Maxn];
map<pair<int,int>,int>rev;
int find(int x){
    if(x==fa[x])return x;
    return fa[x]=find(fa[x]);
}
int son[maxn],size[maxn],dep[maxn],f[maxn];
inline void dfs1(int x,int anc){
    dep [x] = dep [ANC] + 1 ;
    size[x]=1;
    f[x]=anc;
    for(int i=beg[x];i;i=nex[i]){
        int t=to[i];
        if(t==anc)continue;
        dfs1(t,x);
        size[x]+=size[t];
        if(size[t]>size[son[x]])
            son[x]=t;
    }
}
int id[maxn],cnt,top[maxn];
inline void dfs2(int x,int topc){
    id[x]=++cnt;
    top[x]=topc;
    if(!son[x])return;
    dfs2(son[x],topc);
    for(int i=beg[x];i;i=nex[i]){
        int t=to[i];
        if(t==f[x]||t==son[x])
            continue;
        dfs2(t,t); 
    }
}
int tr[maxn],lazy[maxn];
inline void build(int h,int l,int r){
    tr[h]=r-l+1;
    if(l==r)return;
    int mid=(l+r)>>1;
    build(h<<1,l,mid);
    build(h<<1|1,mid+1,r);
}
inline void pushup(int h){
    p [h] = 0 ;
    lazy[h]=1;
}
inline void pushdown(int h){
    if(!lazy[h])return;
    pushup(h<<1);
    pushup(h<<1|1);
    lazy[h]=0;
}
inline void modify(int h,int l,int r,int x,int y){
    if(l>y||r<x)return;
    if(tr[h]==0)return;
    if(l>=x&&r<=y){
        //pushup(h);
        tr[h]=0;
        return;
    }
    //pushdown(h);
    int mid=(l+r)>>1;
    modify(h<<1,l,mid,x,y);
    modify(h<<1|1,mid+1,r,x,y);
    tr[h]=tr[h<<1]+tr[h<<1|1];
}
inline void mc(int x,int y){
    while(top[x]!=top[y]){
        if(dep[top[x]]<dep[top[y]])swap(x,y);
        modify(1,1,n,id[top[x]],id[x]);
        x=f[top[x]];
    }
    if(dep[x]>dep[y])swap(x,y);
    if(x==y)return;
    modify(1,1,n,id[x]+1,id[y]);
}
inline int query(int h,int l,int r,int x,int y){
    if(l>y||r<x)return 0;
    if(tr[h]==0)return 0;
    if(l>=x&&r<=y)return tr[h];
    int mid=(l+r)>>1;
    return query(h<<1,l,mid,x,y)+query(h<<1|1,mid+1,r,x,y);
}
inline int qc(int x,int y){
    int ans=0;
    while(top[x]!=top[y]){
        if(dep[top[x]]<dep[top[y]])swap(x,y);
        ans+=query(1,1,n,id[top[x]],id[x]);
        x=f[top[x]]; 
    }
    if(dep[x]>dep[y])swap(x,y);
    if(x==y)return ans;
    ans+=query(1,1,n,id[x]+1,id[y]);
    return ans;
}
int num, years [maxn], tot;
int main () {
    cin>>n>>m;
    for(int i=1;i<=m;i++){
        scanf("%d%d",&a[i].x,&a[i].y);
        a[i].flag=1;
        rev[make_pair(a[i].x,a[i].y)]=i;
        rev[make_pair(a[i].y,a[i].x)]=i;
    }
    int opt,x,y;
    while(1){
        scanf("%d",&opt);
        if(opt==-1)break;
        scanf("%d%d",&x,&y);
        if(!opt)a[rev[make_pair(x,y)]].flag=0;
        q[++num]=(node){x,y,opt};
    }
    for(int i=1;i<=n;i++)
        fa[i]=i;
    int cnt=0;
    for(int i=1;i<=m;i++){
        if(a[i].flag==0)continue;
        if(find(a[i].x)!=find(a[i].y)){
            fa[find(a[i].x)]=find(a[i].y);
            add(a[i].x,a[i].y);
            add(a[i].y,a[i].x);
            a[i].flag=0;
        }
    }
    dfs1 ( 1 , 0 );
    dfs2(1,1);
    build(1,1,n);
    modify(1,1,n,1,1);
    for(int i=1;i<=m;i++){
        if(!a[i].flag)continue;
        mc(a[i].x,a[i].y);
    }
    for(int i=num;i;i--){
        if(q[i].flag)ans[++tot]=qc(q[i].x,q[i].y);
        else mc(q[i].x,q[i].y);
    }
    for(int i=tot;i;i--)
        printf("%d\n",ans[i]);
    return 0;
}

Deeply felt small and weak.