• CF704E. Iron Man


    题目大意

    一棵树上有m个人,每个人有出现时间、速度以及起点终点,到达终点时会瞬间消失,求最早两人相遇的时间

    n,m<=1e5

    题解

    树剖变成若干线段求交,按照时间排序后set维护即可,相交的两个只可能相邻所以只用考虑相邻两个的

    注意考虑k相等的情况

    code

    #include <bits/stdc++.h>
    #define fo(a,b,c) for (a=b; a<=c; a++)
    #define fd(a,b,c) for (a=b; a>=c; a--)
    #define abs(x) ((x)>0?(x):-(x))
    #define min(a,b) (a<b?a:b)
    #define max(a,b) (a>b?a:b)
    #define E 0.00000001
    #define ll long long
    //#define file
    using namespace std;
    
    struct Type{double x,X,k,b;int s;} c[300001];
    vector<Type> b[100001];
    int a[200001][2],ls[100001],d[100001],fa[100001][17],top[100001],nx[100001],sz[100001],n,m,i,j,k,l,len,x,y,Lca,tot;
    double T,C,ans,sum,s1,s2,tt;
    struct type{double x1,x2,k,b;};
    bool operator < (type a,type b) {return a.k*tt+a.b<b.k*tt+b.b;}
    set<type> st;
    set<type> :: iterator I1,I2;
    
    bool cmp(Type a,Type b) {return abs(a.x-b.x)>E && a.x<b.x || abs(a.x-b.x)<=E && a.s>b.s;}
    void New(int x,int y) {++len;a[len][0]=y;a[len][1]=ls[x];ls[x]=len;}
    void dfs(int Fa,int t)
    {
    	int i,mx=0;
    	sz[t]=1,d[t]=d[Fa]+1;
    	fa[t][0]=Fa;
    	fo(i,1,16) fa[t][i]=fa[fa[t][i-1]][i-1];
    	
    	for (i=ls[t]; i; i=a[i][1])
    	if (a[i][0]!=Fa)
    	{
    		dfs(t,a[i][0]);
    		sz[t]+=sz[a[i][0]];
    		
    		if (sz[a[i][0]]>mx)
    		mx=sz[a[i][0]],nx[t]=a[i][0];
    	}
    }
    void Dfs(int Fa,int t)
    {
    	int i;
    	if (nx[t])
    	top[nx[t]]=top[t],Dfs(t,nx[t]);
    	
    	for (i=ls[t]; i; i=a[i][1])
    	if (a[i][0]!=Fa && a[i][0]!=nx[t])
    	top[a[i][0]]=a[i][0],Dfs(t,a[i][0]);
    }
    
    void swap(int &x,int &y) {int z=x;x=y;y=z;}
    int lca(int x,int y)
    {
    	int i;
    	if (d[x]<d[y]) swap(x,y);
    	fd(i,16,0) if (d[fa[x][i]]>=d[y]) x=fa[x][i];
    	fd(i,16,0) if (fa[x][i]!=fa[y][i]) x=fa[x][i],y=fa[y][i];
    	if (x!=y) x=fa[x][0];
    	return x;
    }
    void add(int t,double X1,double X2,double Y1,double Y2)
    {
    	if (abs(X1-X2)<=E)
    	{
    		b[t].push_back({X1,X1,0,Y1,1});
    		b[t].push_back({X1,X1,0,Y1,-1});
    		return;
    	}
    	double k=(Y2-Y1)/(X2-X1);
    	b[t].push_back({X1,X2,k,Y1-k*X1,1});
    	b[t].push_back({X2,X1,k,Y1-k*X1,-1});
    }
    
    void js(type a,type b)
    {
    	double x,X1,X2;
    	if (abs(a.k-b.k)<=E)
    	{
    		if (abs(a.b-b.b)<=E)
    		{
    			X1=max(a.x1,b.x1),X2=min(a.x2,b.x2);
    			if (abs(X1-X2)<=E || X1<X2) ans=min(ans,X1);
    		}
    		return;
    	}
    	x=(a.b-b.b)/(b.k-a.k),X1=max(a.x1,b.x1),X2=min(a.x2,b.x2);
    	if ((abs(x-X1)<=E || X1<x) && (abs(x-X2)<=E || x<X2)) ans=min(ans,x);
    }
    
    void work(int t)
    {
    	type s1,s2,C;
    	int i,j,k,l;
    	
    	tot=0,l=b[t].size();
    	fo(i,0,l-1) c[++tot]=b[t][i];
    	sort(c+1,c+tot+1,cmp);
    	
    	tt=c[1].x;
    	st.clear();
    	fo(i,1,tot)
    	if (abs(c[i].x-ans)>E && c[i].x<ans)
    	{
    		tt=c[i].x,I2=st.upper_bound({0,0,c[i].k,c[i].b});
    		if (c[i].s==1)
    		{
    			C={c[i].x,c[i].X,c[i].k,c[i].b};
    			if (I2!=st.end())
    			s2=*I2,js(C,s2);
    			if (I2!=st.begin())
    			--I2,s2=*I2,js(C,s2);
    			st.insert(C);
    		}
    		else
    		{
    			C={c[i].X,c[i].x,c[i].k,c[i].b};
    			if (I2!=st.begin() && I2!=st.end())
    			I1=I2,--I1,s1=*I1,s2=*I2,js(s1,s2);
    			st.erase(C);
    		}
    	}
    	else break;
    }
    
    int main()
    {
    	#ifdef file
    	freopen("CF704E.in","r",stdin);
    	#endif
    	
    	scanf("%d%d",&n,&m),ans=2147483647;
    	fo(i,1,n-1) scanf("%d%d",&j,&k),New(j,k),New(k,j);
    	dfs(0,1),top[1]=1,Dfs(0,1);
    	
    	fo(i,1,m)
    	{
    		scanf("%lf%lf%d%d",&T,&C,&x,&y),Lca=lca(x,y),sum=d[x]+d[y]-d[Lca]*2,s1=s2=0;
    		while (1)
    		{
    			if (top[x]==top[y])
    			{
    				if (x==y) add(top[x],T+s1/C,T+s1/C,d[x],d[x]);
    				else
    				add(top[x],T+s1/C,T+(sum-s2)/C,d[x],d[y]);
    				break;
    			}
    			else
    			if (d[top[x]]>d[top[y]])
    			j=fa[top[x]][0],add(top[x],T+s1/C,T+(s1+(d[x]-d[j]))/C,d[x],d[j]),s1+=d[x]-d[j],x=j;
    			else
    			j=fa[top[y]][0],add(top[y],T+(sum-(s2+(d[y]-d[j])))/C,T+(sum-s2)/C,d[j],d[y]),s2+=d[y]-d[j],y=j;
    		}
    	}
    	
    	fo(i,1,n)
    	if (!b[i].empty())
    	work(i);
    	
    	if (ans>1000000000) printf("-1
    ");
    	else printf("%.20lf
    ",ans);
    	
    	fclose(stdin);
    	fclose(stdout);
    	return 0;
    }
    
  • 相关阅读:
    ol之天地图加载
    ol之地图定位
    ol之filter过滤获取feature
    android -- The import android.support cannot be resolved
    docker 笔记(3) -- 容器
    docker 笔记(2) -- 镜像
    docker 笔记(2) -- 镜像
    docker 笔记(0) -- 概述
    docker 笔记(1) -- 安装
    py3 与 Django 连接数据库: No module named 'MySQLdb'
  • 原文地址:https://www.cnblogs.com/gmh77/p/13867140.html
Copyright © 2020-2023  润新知