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

Posted 2022-10-16 --hpy-7m

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。

  1 #define ios ios_base::sync_with_stdio(0); cin.tie(0);
  2 #include <cstdio>//sprintf islower isupper
  3 #include <cstdlib>//malloc  exit strcat itoa system("cls")
  4 #include <iostream>//pair
  5 #include <fstream>//freopen("C:\\Users\\13606\\Desktop\\草稿.txt","r",stdin);
  6 #include <bitset>
  7 //#include <map>
  8 //#include<unordered_map>
  9 #include <vector>
 10 #include <stack>
 11 #include <set>
 12 #include <string.h>//strstr substr
 13 #include <string>
 14 #include <time.h>//srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9;
 15 #include <cmath>
 16 #include <deque>
 17 #include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less
 18 #include <vector>//emplace_back
 19 //#include <math.h>
 20 //#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor
 21 #include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare)
 22 using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation
 23 #define fo(a,b,c) for(register int a=b;a<=c;++a)
 24 #define fr(a,b,c) for(register int a=b;a>=c;--a)
 25 #define mem(a,b) memset(a,b,sizeof(a))
 26 #define pr printf
 27 #define sc scanf
 28 #define ls rt<<1
 29 #define rs rt<<1|1
 30 typedef long long ll;
 31 #define RG register int;
 32 void swapp(int &a,int &b);
 33 double fabss(double a);
 34 int maxx(int a,int b);
 35 int minn(int a,int b);
 36 int Del_bit_1(int n);
 37 int lowbit(int n);
 38 int abss(int a);
 39 //const long long INF=(1LL<<60);
 40 const double E=2.718281828;
 41 const double PI=acos(-1.0);
 42 const int inf=(1<<30);
 43 const double ESP=1e-20;
 44 const int mod=(int)1e9+7;
 45 const int N=(int)1e6+10;
 46 
 47 struct node
 48 
 49     int k,b,id;
 50     long double pos;
 51     friend bool operator<(node a,node b)
 52     
 53         return a.pos<b.pos;
 54     
 55 a[N];
 56 
 57 long double get(int k,int b)
 58 
 59     long double len=1.0*b/(1.0*k);
 60     len=abs(len);
 61     if(k>0)
 62     
 63         if(b>0)
 64             return -len;
 65         else
 66             return len;
 67     
 68     else
 69     
 70         if(b>0)
 71             return len;
 72         else
 73             return -len;
 74     
 75 
 76 bool same(long double x,long double y)
 77 
 78     return abs(x-y)<ESP;
 79 
 80 
 81 ll dp[N],dp2[N];
 82 
 83 int main()
 84 
 85     int n,cnt=0;
 86     long double xx;
 87     sc("%d",&n);
 88     for(int i=1;i<=n;++i)
 89     
 90         int k,b;
 91         sc("%d%d",&k,&b);
 92         if(k==0)
 93             continue;
 94         a[++cnt]=k,b,a[cnt].pos=get(a[cnt].k,a[cnt].b);
 95     
 96     n=cnt;
 97     sort(a+1,a+1+n);
 98     a[1].id=1;
 99     int id=1;
100     for(int i=2;i<=n;++i)
101     
102         if(!same(a[i].pos,a[i-1].pos))
103             id++;
104         a[i].id=id;
105     
106     for(int i=1;i<=n;++i)
107         if(a[i].k>0)
108             dp[a[i].id]+=a[i].k;
109     for(int i=1;i<=n;++i)
110         dp[i]+=dp[i-1];
111     for(int i=n;i>=1;--i)
112     
113         if(a[i].k<0)
114             dp2[a[i].id-1]+=a[i].k;
115     
116     for(int i=n;i>=0;--i)
117         dp2[i]+=dp2[i+1];
118     int ans=0;
119     for(int i=1;i<=id;++i)
120         if(dp[i]+dp2[i]!=dp[i-1]+dp2[i-1])
121             ans++;
122     pr("%d\n",ans);
123     return 0;
124 
125 
126 /**************************************************************************************/
127 
128 int maxx(int a,int b)
129 
130     return a>b?a:b;
131 
132 
133 void swapp(int &a,int &b)
134 
135     a^=b^=a^=b;
136 
137 
138 int lowbit(int n)
139 
140     return n&(-n);
141 
142 
143 int Del_bit_1(int n)
144 
145     return n&(n-1);
146 
147 
148 int abss(int a)
149 
150     return a>0?a:-a;
151 
152 
153 double fabss(double a)
154 
155     return a>0?a:-a;
156 
157 
158 int minn(int a,int b)
159 
160     return a<b?a:b;
161