zoukankan      html  css  js  c++  java
  • [BZOJ4541][HNOI2016]矿区(平面图转对偶图)

    https://www.cnblogs.com/ljh2000-jump/p/6423399.html

     1 #include<cmath>
     2 #include<vector>
     3 #include<cstdio>
     4 #include<algorithm>
     5 #define rep(i,l,r) for (int i=(l); i<=(r); i++)
     6 typedef long long ll;
     7 using namespace std;
     8 
     9 const int N=1200010;
    10 int n,m,Q,x,y,d,cnt=1,nxt[N],tot,rt,bel[N],fa[N],ask[N];
    11 bool vis[N],in[N];
    12 ll s[N],sp[N],fz,fm;
    13 
    14 struct P{ int x,y; }p[N];
    15 P operator -(const P &a,const P &b){ return (P){a.x-b.x,a.y-b.y}; }
    16 ll operator *(const P &a,const P &b){ return 1ll*a.x*b.y-1ll*b.x*a.y; }
    17 
    18 struct E{
    19     int u,v,id; double sl;
    20     E(){}; E(int a,int b,int i):u(a),v(b),id(i),sl(atan2(p[b].y-p[a].y,p[b].x-p[a].x)){}
    21 }e[N];
    22 bool operator <(const E &a,const E &b){ return a.sl<b.sl; }
    23 vector<E>w[N],tr[N];
    24 void add(int u,int v){ cnt++; e[cnt]=(E){u,v,cnt}; w[u].push_back(e[cnt]); }
    25 
    26 ll gcd(ll a,ll b){ return b ? gcd(b,a%b) : a; }
    27 
    28 int find(int x,E b){
    29     int l=0,r=w[x].size()-1;
    30     while (l<r){
    31         int mid=(l+r+1)>>1;
    32         if (b<w[x][mid]) r=mid-1; else l=mid;
    33     }
    34     return l;
    35 }
    36 
    37 void dfs(int x){
    38     vis[x]=1; sp[x]=s[x]*s[x]; s[x]<<=1; int ed=tr[x].size()-1;
    39     rep(i,0,ed){
    40         int v=tr[x][i].v; if (vis[v]) continue;
    41         fa[v]=x; in[tr[x][i].id]=in[tr[x][i].id^1]=1;
    42         dfs(v); s[x]+=s[v]; sp[x]+=sp[v];
    43     }
    44 }
    45 
    46 int main(){
    47     freopen("bzoj4541.in","r",stdin);
    48     freopen("bzoj4541.out","w",stdout);
    49     scanf("%d%d%d",&n,&m,&Q);
    50     rep(i,1,n) scanf("%d%d",&x,&y),p[i]=(P){x,y};
    51     rep(i,1,m) scanf("%d%d",&x,&y),add(x,y),add(y,x);
    52     rep(i,1,n) sort(w[i].begin(),w[i].end());
    53     rep(i,2,cnt){
    54         int ne=find(e[i].v,e[i^1])-1;
    55         if (ne==-1) ne=w[e[i].v].size()-1;
    56         nxt[i]=w[e[i].v][ne].id;
    57     }
    58     rep(i,2,cnt) if (!bel[i]){
    59         bel[i]=bel[nxt[i]]=++tot;
    60         for (int x=nxt[i]; e[x].v!=e[i].u; x=nxt[x],bel[x]=tot)
    61             s[tot]+=(p[e[x].u]-p[e[i].u])*(p[e[x].v]-p[e[i].u]);
    62         if (s[tot]<=0) rt=tot;
    63     }
    64     rep(i,2,cnt) tr[bel[i]].push_back(E(bel[i],bel[i^1],i));
    65     dfs(rt);
    66     while (Q--){
    67         scanf("%d",&d); d=(d+fz)%n+1;
    68         rep(i,1,d) scanf("%d",&ask[i]),ask[i]=(ask[i]+fz)%n+1;
    69         ask[d+1]=ask[1]; fz=fm=0;
    70         rep(i,1,d){
    71             int x=w[ask[i]][find(ask[i],E(ask[i],ask[i+1],0))].id;
    72             if (!in[x]) continue;
    73             if (fa[bel[x]]==bel[x^1]) fm+=s[bel[x]],fz+=sp[bel[x]];
    74                 else fm-=s[bel[x^1]],fz-=sp[bel[x^1]];
    75         }
    76         ll d=gcd(fz,fm); fz/=d; fm/=d; printf("%lld %lld
    ",fz,fm);
    77     }
    78     return 0;
    79 }
  • 相关阅读:
    ASP.NET 2.0 中的代码隐藏和编译
    魅族M8 UI新架构
    详细讲解大型数据库的设计原则与开发技巧
    [ZT]精彩的国外育儿教育读本,图文并茂
    企業導入ERP項目要防止「君子」作亂
    [ZT]VS2005 中 Lc.exe已退出,代码为 1,今天剛遇到就找了
    图像词汇
    VS2005 + VSS2005 的方法
    几种SQL Server数据库分页方式
    DB2 Express版与SQL Server标准版技术比较
  • 原文地址:https://www.cnblogs.com/HocRiser/p/10630891.html
Copyright © 2011-2022 走看看