Orz PoPoQQQ (UPD:ydc的写法好像更熟悉一些……(类似堆优化的Dij啊~
先留个坑……明天再看一看……感觉好神奇>_<(完美消除序列之于弦图 就好似 拓扑序列之于DAG,所以弦图的问题许多都要靠这个完美消除序列来做)
1 /************************************************************** 2 Problem: 1006 3 User: Tunix 4 Language: C++ 5 Result: Accepted 6 Time:536 ms 7 Memory:34996 kb 8 ****************************************************************/ 9 10 //BZOJ 1006 11 #include<vector> 12 #include<cstdio> 13 #include<cstdlib> 14 #include<cstring> 15 #include<iostream> 16 #include<algorithm> 17 #define rep(i,n) for(int i=0;i<n;++i) 18 #define F(i,j,n) for(int i=j;i<=n;++i) 19 #define D(i,j,n) for(int i=j;i>=n;--i) 20 using namespace std; 21 22 int getint(){ 23 int v=0,sign=1; char ch=getchar(); 24 while(ch<'0'||ch>'9') {if (ch=='-') sign=-1; ch=getchar();} 25 while(ch>='0'&&ch<='9') {v=v*10+ch-'0'; ch=getchar();} 26 return v*sign; 27 } 28 typedef long long LL; 29 const int N=100010,INF=~0u>>2; 30 /*******************tamplate********************/ 31 struct List{ 32 int to,next; 33 }table[4004004]; 34 int head[N],tot; 35 int n,m,ans,best,f[N],list[N],seq[N],color[N],mark[N]; 36 bool v[N]; 37 void add(int *h,int x,int y){ 38 table[++tot].to=y; 39 table[tot].next=h[x]; 40 h[x]=tot; 41 } 42 void MCS(){ 43 int i,j; 44 F(i,1,n) add(list,0,i); 45 D(j,n,1){ 46 while(1){ 47 for(i=list[best];i;i=table[i].next){ 48 if (!v[table[i].to]) break; 49 else list[best]=table[i].next; 50 } 51 if (i){ 52 int x=table[i].to; 53 v[x]=1; seq[j]=x; 54 for(i=head[x];i;i=table[i].next) 55 if(!v[table[i].to]){ 56 f[table[i].to]++; 57 add(list,f[table[i].to],table[i].to); 58 best=max(best,f[table[i].to]); 59 } 60 break; 61 }else best--; 62 } 63 } 64 } 65 int main(){ 66 #ifndef ONLINE_JUDGE 67 freopen("input.txt","r",stdin); 68 // freopen("output.txt","w",stdout); 69 #endif 70 n=getint(); m=getint(); 71 int x,y; 72 F(i,1,m){ 73 x=getint(); y=getint(); 74 add(head,x,y); 75 add(head,y,x); 76 } 77 MCS(); 78 D(j,n,1){ 79 int x=seq[j],i; 80 for(int i=head[x];i;i=table[i].next) 81 mark[ color[table[i].to] ]=j; 82 for(i=1;i<=n && mark[i]==j;i++); 83 color[x]=i; 84 ans=max(ans,i); 85 } 86 printf("%d\n",ans); 87 return 0; 88 }