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 }
原文地址：https://www.cnblogs.com/--HPY-7m/p/11508712.html
时间： 2024-08-12 02:41:53