CF924D. Contact ATC

Solution

第一波轉(zhuǎn)化挺妙的。

令$t{l}_i$表示用$?w$的風(fēng)到原點(diǎn)的時(shí)間，$t{r}_i$表示用$w$的風(fēng)到圓點(diǎn)的時(shí)間。

顯然$(i,j)$有解的條件為$(t{l}_i?t{l}_j)(t{r}_i?t{r}_j)<=0$。

然后就變成了二維偏序問題（據(jù)說會(huì)卡精度，然后就寫分?jǐn)?shù)類了）。

Code

#include <vector> #include <list> #include <map> #include <set> #include <deque> #include <queue> #include <stack> #include <bitset> #include <algorithm> #include <functional> #include <numeric> #include <utility> #include <sstream> #include <iostream> #include <iomanip> #include <cstdio> #include <cmath> #include <cstdlib> #include <cctype> #include <string> #include <cstring> #include <ctime> #include <cassert> #include <string.h> //#include <unordered_set> //#include <unordered_map> //#include <bits/stdc++.h>#define MP(A,B) make_pair(A,B) #define PB(A) push_back(A) #define SIZE(A) ((int)A.size()) #define LEN(A) ((int)A.length()) #define FOR(i,a,b) for(int i=(a);i<(b);++i) #define fi first #define se second #define int llusing namespace std;template<typename T>inline bool upmin(T &x,T y) { return y<x?x=y,1:0; } template<typename T>inline bool upmax(T &x,T y) { return x<y?x=y,1:0; }typedef long long ll; typedef unsigned long long ull; typedef long double lod; typedef pair<int,int> PR; typedef vector<int> VI;const lod eps=1e-11; const lod pi=acos(-1); const int oo=1<<30; const ll loo=1ll<<62; const int mods=998244353; const int MAXN=200005; const int INF=0x3f3f3f3f;//1061109567 /*--------------------------------------------------------------------*/ inline int read() {int f=1,x=0; char c=getchar();while (c<'0'||c>'9') { if (c=='-') f=-1; c=getchar(); }while (c>='0'&&c<='9') { x=(x<<3)+(x<<1)+(c^48); c=getchar(); }return x*f; } int num=0; struct number { int x,y; } b[MAXN]; bool operator < (number a,number b) { return a.x*b.y<a.y*b.x; } bool operator == (number a,number b) { return a.x*b.y==a.y*b.x; } bool operator > (number a,number b) { return a.x*b.y>a.y*b.x; } pair<number,number> a[MAXN];int s[MAXN<<2],c[MAXN],n,w; int query(int x) { int ans=0; for (;x;x-=x&(-x)) ans+=s[x]; return ans; } void add(int x,int y) { for (;x<=n;x+=x&(-x)) s[x]+=y; } signed main() {n=read(),w=read();for (int i=1;i<=n;i++){int x=read(),y=read();if (y>0) a[i].fi=(number){-x,y-w},a[i].se=(number){-x,y+w},b[i]=a[i].se;else a[i].fi=(number){x,w-y},a[i].se=(number){x,-y-w},b[i]=a[i].se;}sort(b+1,b+n+1); c[0]=1;for (int i=1;i<=n;i++) if (b[i]==b[i-1]) c[i]=c[i-1];else c[i]=c[i-1]+1;sort(a+1,a+n+1,[&](pair<number,number> x,pair<number,number> y){ return (x.fi<y.fi)||(x.fi==y.fi&&x.se>y.se); }); // for (int i=1;i<=n;i++) cout<<1.0*a[i].fi.x/a[i].fi.y<<" "<<1.0*a[i].se.x/a[i].se.y<<endl;int ans=0;for (int i=1;i<=n;i++){int t=c[lower_bound(b+1,b+n+1,a[i].se)-b];ans+=query(n)-query(t-1);add(t,1);}printf("%lld\n",ans);return 0; }

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