codeforces 813F Bipartite Checking

最新推荐文章于 2020-02-11 16:16:14 发布

elijahqi

最新推荐文章于 2020-02-11 16:16:14 发布

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本文提供了一种解决 CodeForces 813F 题目“ Bipartite Checking ”的有效算法。该算法通过维护一棵最大生成树来检查每一次增删边操作后图是否仍为二部图。文章详细介绍了实现过程，包括数据结构设计和关键操作如旋转、查找等。

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http://www.elijahqi.win/2018/03/09/codeforces-813f-bipartite-checking/
F. Bipartite Checking
time limit per test

6 seconds
memory limit per test

256 megabytes
input

standard input
output

standard output

You are given an undirected graph consisting of n vertices. Initially there are no edges in the graph. Also you are given q queries, each query either adds one undirected edge to the graph or removes it. After each query you have to check if the resulting graph is bipartite (that is, you can paint all vertices of the graph into two colors so that there is no edge connecting two vertices of the same color).
Input

The first line contains two integers n and q (2 ≤ n, q ≤ 100000).

Then q lines follow. ith line contains two numbers xi and yi (1 ≤ xi < yi ≤ n). These numbers describe ith query: if there is an edge between vertices xi and yi, then remove it, otherwise add it.
Output

Print q lines. ith line must contain YES if the graph is bipartite after ith query, and NO otherwise.
Example

input
Copy

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

output

YES
YES
NO
YES
NO

同bzoj4025 但我原来的方法不会搞了所以选择学习了下icefox巨佬的方法

还是维护一颗最大生成树树的权值是这条边消失的时间记录mark表示判断该边是否属于奇环或者偶环或者仅仅是树边每次如果新边消失时间更长则替换树上原有的边每次有一个奇环就计数器+1 如果该操作是删除就正常处理注意需要删除树边

#include<map>
#include<queue>
#include<cstdio>
#include<algorithm>
#define pa pair<int,int>
#define N 220000
#define inf 0x3f3f3f3f
using namespace std;
map<pa,int> mm;
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,f=1;char ch=gc();
    while(ch<'0'||ch>'9') {if (ch=='-') f=-1;ch=gc();}
    while(ch<='9'&&ch>='0') x=x*10+ch-'0',ch=gc();
    return x*f;
}
struct node{
    int id,op;
}qr[N];
struct node1{
    int x,y;
}eg[N];
int c[N][2],fa[N],b[N],v[N],size[N],q[N],top,sum,mark[N],n,m,Q;
bool rev[N];
inline bool isroot(int x){
    return c[fa[x]][0]!=x&&c[fa[x]][1]!=x;
}
inline void pushdown(int x){
    if (!rev[x]) return;rev[x]^=1;
    int l=c[x][0],r=c[x][1];rev[l]^=1;
    swap(c[x][0],c[x][1]);rev[r]^=1;
}
inline void update(int x){
    int l=c[x][0],r=c[x][1];b[x]=x;
    size[x]=size[l]+size[r]+1;  
    if (v[b[l]]<v[b[x]]) b[x]=b[l];
    if (v[b[r]]<v[b[x]]) b[x]=b[r];
}
inline void rotate(int x){
    int y=fa[x],z=fa[y],l=c[y][1]==x,r=l^1;
    if(!isroot(y)) c[z][c[z][1]==y]=x;
    fa[c[x][r]]=y;fa[y]=x;fa[x]=z;
    c[y][l]=c[x][r];c[x][r]=y;update(y);update(x);
}
inline void splay(int x){
    q[top=1]=x;for (int i=x;!isroot(i);i=fa[i]) q[++top]=fa[i];
    while(top) pushdown(q[top--]);
    while(!isroot(x)){
        int y=fa[x],z=fa[y];
        if (!isroot(y)){
            if (c[y][0]==x^c[z][0]==y) rotate(x);else rotate(y);
        }rotate(x);
    }
}
inline void access(int x){
    for (int t=0;x;t=x,x=fa[x]) splay(x),c[x][1]=t,update(x);
}
inline int find(int x){
    access(x);splay(x);while(c[x][0]) x=c[x][0];return x;
}
inline void makeroot(int x){
    access(x);splay(x);rev[x]^=1;
}
inline link(int x,int y){makeroot(x);fa[x]=y;}
inline void cut(int x,int y){
    makeroot(x);access(y);splay(y);c[y][0]=fa[c[y][0]]=0;update(y);
}
int main(){
    freopen("cf813f.in","r",stdin);
    n=read();Q=read();int m=0;v[0]=inf;
    for(int i=1;i<=Q;++i){
        int x=read(),y=read();if(x>y) swap(x,y);int id=mm[make_pair(x,y)];
        if(id){qr[i].op=2;qr[i].id=id;mm[make_pair(x,y)]=0;v[n+id]=i;b[n+id]=n+id;continue;}
        qr[i].op=1;qr[i].id=++m;eg[m].x=x;eg[m].y=y;mm[make_pair(x,y)]=m;
    }
    for (int i=1;i<=n;++i) v[i]=inf,b[i]=i;
    for (int i=1;i<=m;++i) if (!v[i+n]) {v[i+n]=Q+1;b[i+n]=i+n;}
    for (int i=1;i<=Q;++i){
        int id=qr[i].id;int x=eg[id].x,y=eg[id].y;
        if (qr[i].op==1){
            if (find(y)!=find(x)) link(x,id+n),link(y,id+n);else{
                makeroot(x);access(y);splay(y);int tt=b[y],sz=size[y];
                if ((sz-1>>1)&1){
                    if (v[id+n]>v[tt]) cut(eg[tt-n].x,tt),cut(eg[tt-n].y,tt),link(x,id+n),link(y,id+n),mark[tt-n]=2;else mark[id]=2;
                }else{
                    ++sum;
                    if (v[id+n]>v[tt]) cut(eg[tt-n].x,tt),cut(eg[tt-n].y,tt),link(x,id+n),link(y,id+n),mark[tt-n]=1;else mark[id]=1;
                }
            }
        }else{if (!mark[id]) cut(x,id+n),cut(y,id+n);if (mark[id]==1) --sum;}
        if (sum) puts("NO");else puts("YES");
    }
    return 0;
}