codeforces 813F Bipartite Checking
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
#includeint> 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]]x]]) b[x]=b[l];
if (v[b[r]]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;
}