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  • 并不对劲的loj2179:p3714:[BJOI2017]树的难题

    题目大意

    有一棵树,(n)((nleq2*10^5))个点,每条边(i)有颜色(w_i),共有(m)((mleq n))种颜色,第(i)种颜色的权值是(c_i)((|c_i|leq10^4))
    定义一条路径的权值是该路径上所有同色段的颜色的权值之和
    给定(l,r),求边数在([l,r])中权值最大的路径的权值

    题解

    将每个点的所有边按颜色排序后,对这棵树进行点分治,每次统计过当前重心的路径
    用线段树统计应该挺板的吧
    有人用单调队列做,然而我不会,先坑着

    代码
    #include<algorithm>
    #include<cmath>
    #include<cstdio>
    #include<cstdlib>
    #include<cstring>
    #include<ctime>
    #include<iomanip>
    #include<iostream>
    #include<map>
    #include<queue>
    #include<set>
    #include<stack>
    #include<vector>
    #define rep(i,x,y) for(register int i=(x);i<=(y);++i)
    #define dwn(i,x,y) for(register int i=(x);i>=(y);--i)
    #define pii pair<int ,int>
    #define fi first
    #define se second
    #define mp make_pair
    #define pb push_back
    #define maxn 200010 
    #define inf 2147483647
    #define ls (u<<1)
    #define rs (u<<1|1)
    #define mi (L+R>>1)
    #define vv(x) v[x][i].se
    using namespace std;
    int read()
    {
    	int x=0,f=1;char ch=getchar();
    	while(!isdigit(ch)&&ch!='-')ch=getchar();
    	if(ch=='-')f=-1,ch=getchar();
    	while(isdigit(ch))x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
    	return x*f;
    }
    void write(int x)
    {
    	if(x==0){putchar('0'),putchar('
    ');return;}
    	int f=0;char ch[20];
    	if(x<0)putchar('-'),x=-x;
    	while(x)ch[++f]=x%10+'0',x/=10;
    	while(f)putchar(ch[f--]);
    	putchar('
    ');
    	return;
    }
    int n,m,l,r,ans=-inf,wt,sumsiz,mnsz,vis[maxn];
    long long cnt;
    int siz[maxn],c[maxn];
    vector<pii >v[maxn];
    struct tree
    {
    	int tr[maxn<<2],mk[maxn<<2];
    	void pu(int u){tr[u]=max(tr[ls],tr[rs]);}
    	void mark(int u,int k){mk[u]=tr[u]=k;}
    	void pd(int u){if(mk[u]){mark(ls,mk[u]),mark(rs,mk[u]),mk[u]=0;}}
    	void add(int u,int L,int R,int x,int k)
    	{
    		if(x<=L&&R<=x){tr[u]=max(k,tr[u]),mk[u]=0;return;}
    		pd(u);
    		if(x<=mi)add(ls,L,mi,x,k);
    		else add(rs,mi+1,R,x,k);
    		return pu(u);
    	}
    	int ask(int u,int L,int R,int x,int y)
    	{
    		if(y<L||R<x)return -inf;
    		if(x<=L&&R<=y)return tr[u];
    		pd(u);
    		int res=-inf;
    		if(x<=mi)res=ask(ls,L,mi,x,y);
    		if(y>mi)res=max(res,ask(rs,mi+1,R,x,y));
    		return res;
    	}
    }t[2];
    void getwt(int u,int fa)
    {
    	siz[u]=1;int nwmx=0,lim=v[u].size();
    	rep(i,0,lim-1)if(vv(u)!=fa&&!vis[vv(u)])getwt(vv(u),u),siz[u]+=siz[vv(u)],nwmx=max(nwmx,siz[vv(u)]);
    	nwmx=max(nwmx,sumsiz-siz[u]);
    	if(nwmx<mnsz)wt=u,mnsz=nwmx;
    	return;
    }
    void asktr(int u,int fa,int fac,int dep,int num)
    {
    	if(dep>r)return;
    	int lim=v[u].size(),tmp=max(t[0].ask(1,0,n,l-dep,r-dep),t[1].ask(1,0,n,l-dep,r-dep));if(tmp!=-inf)ans=max(ans,tmp+num);
    	rep(i,0,lim-1)if(!vis[vv(u)]&&vv(u)!=fa)asktr(vv(u),u,v[u][i].fi,dep+1,num+(v[u][i].fi==fac?0:c[v[u][i].fi]));
    }
    void addtr(int u,int fa,int fac,int dep,int num,int f)
    {
    	if(dep>r)return;
    	int lim=v[u].size();t[f].add(1,0,n,dep,num);
    	rep(i,0,lim-1)if(!vis[vv(u)]&&vv(u)!=fa)addtr(vv(u),u,v[u][i].fi,dep+1,num+(v[u][i].fi==fac?0:c[v[u][i].fi]),f);
    }
    void getans(int u,int nowsiz)
    {
    	sumsiz=nowsiz,mnsz=n+1,getwt(u,0);int now=wt;
    	int lim=v[now].size(),p=0;t[0].mark(1,-inf),t[1].mark(1,-inf);t[0].add(1,0,n,0,0);
    	cnt+=(long long)nowsiz;
    	rep(i,0,lim-1)
    	{
    		if(!vis[vv(now)])
    		{
    			asktr(vv(now),now,v[now][i].fi,1,c[v[now][i].fi]),
    			addtr(vv(now),now,v[now][i].fi,1,0,1);
    		}
    		if(i!=lim-1&&v[now][i].fi!=v[now][i+1].fi)
    		{
    			rep(j,p,i)if(!vis[v[now][j].se])addtr(v[now][j].se,now,v[now][j].fi,1,c[v[now][j].fi],0);
    			t[1].mark(1,-inf);
    			p=i+1;
    		}
    	}
    	vis[now]=1;
    	rep(i,0,lim-1)if(!vis[vv(now)])getans(vv(now),siz[vv(now)]>siz[now]?nowsiz-siz[now]:siz[vv(now)]);
    	return;
    }
    int main()
    {
    	n=read(),m=read(),l=read(),r=read();if(l<=0&&0<=r)ans=0;
    	rep(i,1,m)c[i]=read();
    	rep(i,1,n-1){int x=read(),y=read(),z=read();v[x].pb(mp(z,y)),v[y].pb(mp(z,x));}
    	rep(i,1,n)sort(v[i].begin(),v[i].end());
    	getans(1,n);
    	write(ans);
    	return 0;
    }
    /*
    5 3 1 4
    -1 -5 -2
    1 2 1
    1 3 1
    2 4 2
    2 5 3
    */
    /*
    8 4 3 4
    -7 9 6 1
    1 2 1
    1 3 2
    1 4 1
    2 5 1
    5 6 2
    3 7 1
    3 8 3
    */
    
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  • 原文地址:https://www.cnblogs.com/xzyf/p/10455146.html
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