Analyzing Polyline -- Codeforces Round #123 (Div. 2)

题意：https://codeforc.es/problemset/problem/195/D

求折线段数。

思路：

对pos进行sort，对不同区间段加k，两个dp处理不同k>0 or k<0前后缀，判断即可。

注意：long double，ESP=1e-20。

 #define IOS ios_base::sync_with_stdio(0); cin.tie(0);
 #include <cstdio>//sprintf islower isupper
 #include <cstdlib>//malloc  exit strcat itoa system("cls")
 #include <iostream>//pair
 #include <fstream>//freopen("C:\\Users\\13606\\Desktop\\草稿.txt","r",stdin);
 #include <bitset>
 //#include <map>
 //#include<unordered_map>
 #include <vector>
 #include <stack>
 #include <set>
 #include <string.h>//strstr substr
 #include <string>
 #include <time.h>//srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9;
 #include <cmath>
 #include <deque>
 #include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less
 #include <vector>//emplace_back
 //#include <math.h>
 //#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor
 #include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare)
 using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation
 #define fo(a,b,c) for(register int a=b;a<=c;++a)
 #define fr(a,b,c) for(register int a=b;a>=c;--a)
 #define mem(a,b) memset(a,b,sizeof(a))
 #define pr printf
 #define sc scanf
 #define ls rt<<1
 #define rs rt<<1|1
 typedef long long ll;
 #define RG register int;
 void swapp(int &a,int &b);
 double fabss(double a);
 int maxx(int a,int b);
 int minn(int a,int b);
 int Del_bit_1(int n);
 int lowbit(int n);
 int abss(int a);
 //const long long INF=(1LL<<60);
 const double E=2.718281828;
 const double PI=acos(-1.0);
 const int inf=(<<);
 const double ESP=1e-;
 const int mod=(int)1e9+;
 const int N=(int)1e6+;

 struct node
 {
     int k,b,id;
     long double pos;
     friend bool operator<(node a,node b)
     {
         return a.pos<b.pos;
     }
 }a[N];

 long double get(int k,int b)
 {
     long double len=1.0*b/(1.0*k);
     len=abs(len);
     if(k>)
     {
         if(b>)
             return -len;
         else
             return len;
     }
     else
     {
         if(b>)
             return len;
         else
             return -len;
     }
 }
 bool same(long double x,long double y)
 {
     return abs(x-y)<ESP;
 }

 ll dp[N],dp2[N];

 int main()
 {
     int n,cnt=;
     long double xx;
     sc("%d",&n);
     for(int i=;i<=n;++i)
     {
         int k,b;
         sc("%d%d",&k,&b);
         if(k==)
             continue;
         a[++cnt]={k,b},a[cnt].pos=get(a[cnt].k,a[cnt].b);
     }
     n=cnt;
     sort(a+,a++n);
     a[].id=;
     int id=;
     for(int i=;i<=n;++i)
     {
         if(!same(a[i].pos,a[i-].pos))
             id++;
         a[i].id=id;
     }
     for(int i=;i<=n;++i)
         if(a[i].k>)
             dp[a[i].id]+=a[i].k;
     for(int i=;i<=n;++i)
         dp[i]+=dp[i-];
     for(int i=n;i>=;--i)
     {
         if(a[i].k<)
             dp2[a[i].id-]+=a[i].k;
     }
     for(int i=n;i>=;--i)
         dp2[i]+=dp2[i+];
     int ans=;
     for(int i=;i<=id;++i)
         if(dp[i]+dp2[i]!=dp[i-]+dp2[i-])
             ans++;
     pr("%d\n",ans);
     return ;
 }

 /**************************************************************************************/

 int maxx(int a,int b)
 {
     return a>b?a:b;
 }

 void swapp(int &a,int &b)
 {
     a^=b^=a^=b;
 }

 int lowbit(int n)
 {
     return n&(-n);
 }

 int Del_bit_1(int n)
 {
     return n&(n-);
 }

 int abss(int a)
 {
     return a>?a:-a;
 }

 double fabss(double a)
 {
     return a>?a:-a;
 }

 int minn(int a,int b)
 {
     return a<b?a:b;
 }