• 【比赛】NOIP2018 旅行


    发现 (m) 只有两种取值,于是可做了

    树的直接贪心

    图的枚举环上的边去掉,然后做树的贪心,搜的时候剪一下枝吧

    写得有点乱

    #include<bits/stdc++.h>
    #define ui unsigned int
    #define ll long long
    #define db double
    #define ld long double
    #define ull unsigned long long
    #define ft first
    #define sd second
    #define pb(a) push_back(a)
    #define mp(a,b) std::make_pair(a,b)
    #define REP(a,b,c) for(register int a=(b),a##end=(c);a<=a##end;++a)
    #define DEP(a,b,c) for(register int a=(b),a##end=(c);a>=a##end;--a)
    const int MAXN=5000+10;
    int n,m,e=1,key,beg[MAXN],nex[MAXN<<1],to[MAXN<<1],was[MAXN<<1],cnt,dfn[MAXN],ans[MAXN],vis[MAXN],arv[MAXN];
    std::vector<int> V[MAXN];
    std::pair<int,int> ext,from[MAXN];
    struct Unique{
    	int fa[MAXN];
    	inline void init(int x)
    	{
    		REP(i,1,x)fa[i]=i;
    	}
    	inline int found(int x)
    	{
    		if(fa[x]!=x)fa[x]=found(fa[x]);
    		return fa[x];
    	}
    	inline int link(int x,int y,int val)
    	{
    		int u=found(x),v=found(y),res=0;
    		if(u!=v)fa[u]=v,res=val;
    		return res;
    	}
    };
    Unique U;
    template<typename T> inline void read(T &x)
    {
    	T data=0,w=1;
    	char ch=0;
    	while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
    	if(ch=='-')w=-1,ch=getchar();
    	while(ch>='0'&&ch<='9')data=((T)data<<3)+((T)data<<1)+(ch^'0'),ch=getchar();
    	x=data*w;
    }
    template<typename T> inline void write(T x,char ch='')
    {
    	if(x<0)putchar('-'),x=-x;
    	if(x>9)write(x/10);
    	putchar(x%10+'0');
    	if(ch!='')putchar(ch);
    }
    template<typename T> inline bool chkmin(T &x,T y){return y<x?(x=y,true):false;}
    template<typename T> inline bool chkmax(T &x,T y){return y>x?(x=y,true):false;}
    template<typename T> inline T min(T x,T y){return x<y?x:y;}
    template<typename T> inline T max(T x,T y){return x>y?x:y;}
    inline void insert(int x,int y,int z)
    {
    	to[++e]=y;
    	nex[e]=beg[x];
    	beg[x]=e;
    	was[e]=z;
    }
    inline void sortdfs(int x,int f)
    {
    	if(arv[x])return ;arv[x]=1;
    	for(register int i=beg[x];i;i=nex[i])V[x].pb(to[i]),sortdfs(to[i],x);
    	std::sort(V[x].begin(),V[x].end());
    }
    inline void initdfs(int x,int f)
    {
    	for(register int i=beg[x];i;i=nex[i])
    		if(to[i]==f||!was[i])continue;
    		else initdfs(to[i],x),from[to[i]]=mp(x,i);
    }
    inline bool dfs(int x,int f,int &fl)
    {
    	dfn[++cnt]=x;
    	if(dfn[cnt]<ans[cnt])chkmin(fl,0);
    	if(fl&&dfn[cnt]>ans[cnt])return false;
    	int lack=-1;
    	for(register int i=beg[x];i;i=nex[i])
    		if(to[i]==f)continue;
    		else if(!was[i])lack=to[i];
    	REP(i,0,V[x].size()-1)
    		if(V[x][i]==lack||V[x][i]==f)continue;
    		else if(!dfs(V[x][i],x,fl))return false;
    	return true;
    }
    inline int LCA(int u,int v)
    {
    	while(u)vis[u]=1,u=from[u].ft;
    	while(v)
    		if(vis[v])return v;
    		else v=from[v].ft;
    }
    inline void solve(int x)
    {
    	int fl;
    	while(x!=key)
    	{
    		was[from[x].sd]=was[from[x].sd^1]=0;
    		cnt=0;if(dfs(1,0,fl=1))memcpy(ans,dfn,sizeof(ans));
    		was[from[x].sd]=was[from[x].sd^1]=1;
    		x=from[x].ft;
    	}
    }
    int main()
    {
    	freopen("travel.in","r",stdin);
    	freopen("travel.out","w",stdout);
    	read(n);read(m);
    	U.init(n);int fl;
    	REP(i,1,n)ans[i]=n+1;
    	REP(i,1,m)
    	{
    		int u,v;read(u);read(v);
    		if(U.link(u,v,1))insert(u,v,1),insert(v,u,1);
    		else ext=mp(u,v);
    	}
    	if(m==n-1)sortdfs(1,0),dfs(1,0,fl=1),memcpy(ans,dfn,sizeof(ans));
    	else
    	{
    		initdfs(1,0);
    		insert(ext.ft,ext.sd,0);
    		insert(ext.sd,ext.ft,0);
    		sortdfs(1,0);
    		dfs(1,0,fl=1),memcpy(ans,dfn,sizeof(ans));
    		was[e]=was[e-1]=1;
    		key=LCA(ext.ft,ext.sd);
    		solve(ext.ft),solve(ext.sd);
    	}
    	REP(i,1,n)write(ans[i],i==iend?'
    ':' ');
    	return 0;
    }
    
  • 相关阅读:
    (二)服务器性能剖析
    (一) MySQL架构
    HBase学习笔记一
    Hadoop系列读书笔记
    Java基础小结
    Redis学习笔记一
    Hive学习笔记一
    使用redis-benchmark测试redis性能
    关闭Stackexchange.Redis的未用到的pub/sub连接
    redis报错:EXCEPTION_ACCESS_VIOLATION
  • 原文地址:https://www.cnblogs.com/hongyj/p/10206214.html
Copyright © 2020-2023  润新知