【链接】点击打开链接
【题意】
n个点,x从左到右严格递增的顺序给出
让你划两条平行的,且没有相同点的直线;
使得它们俩各自最少穿过一个点.
且它们俩穿过了所有的点。
【题解】
枚举第一个点和哪个点组成了一条线,把在线上的点去掉,然后看看剩下的点是不是组成了一条和它平行的线。且穿过了所有的点。
再枚举第一个点单独组成一条线的情况.
O(n^2)复杂度
【错的次数】
0
【反思】
在这了写反思
【代码】
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <map>
#include <queue>
#include <iomanip>
#include <set>
#include <cstdlib>
#include <cmath>
#include <bitset>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb emplace_back
#define fi first
#define se second
#define ld long double
#define ms(x,y) memset(x,y,sizeof x)
#define ri(x) scanf("%d",&x)
#define rl(x) scanf("%lld",&x)
#define rs(x) scanf("%s",x)
#define rf(x) scnaf("%lf",&x)
#define oi(x) printf("%d",x)
#define ol(x) printf("%lld",x)
#define oc putchar(' ')
#define os(x) printf(x)
#define all(x) x.begin(),x.end()
#define Open() freopen("F:\rush.txt","r",stdin)
#define Close() ios::sync_with_stdio(0)
#define sz(x) ((int) x.size())
#define ld long double
typedef pair<int,int> pii;
typedef pair<LL,LL> pll;
//mt19937 myrand(time(0));
//int get_rand(int n){return myrand()%n + 1;}
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int N = 1e3;
struct abc{
LL x,y;
};
int n;
abc a[N+10];
bool bo[N+10];
bool ok(int i,int j,int k){
LL x1 = a[i].x,y1 = a[i].y,x2 = a[j].x,y2 = a[j].y,x3 = a[k].x,y3 = a[k].y;
return (x3-x1)*(y2-y1)==(x2-x1)*(y3-y1);
}
bool ok2(int i,int j,int k,int l){
LL x1 = a[i].x,y1 = a[i].y,x2 = a[j].x,y2 = a[j].y;
LL x3 = a[k].x,y3 = a[k].y,x4 = a[l].x,y4 = a[l].y;
return (y2-y1)*(x4-x3) == (y4-y3)*(x2-x1);
}
int main(){
//Open();
//Close();
ri(n);
rep1(i,1,n){
LL y;
rl(y);
a[i].x = i,a[i].y = y;
}
rep1(i,2,n){
ms(bo,0);
bo[1] = bo[i] = true;
rep1(j,1,n){
if (ok(1,i,j))
bo[j] = true;
}
int fi = -1;
rep1(k,1,n)
if (!bo[k]){
bo[k] = true;
fi = k;
rep1(j,k+1,n)
if (!bo[j]){
if (!ok2(1,i,k,j)) break;
bo[j] = true;
rep1(l,j+1,n)
if (!bo[l] && ok(k,j,l))
bo[l] = true;
break;
}
break;
}
if (fi==-1) continue;
int ok = true;
rep1(j,1,n)
if (!bo[j]){
ok = false;
break;
}
if (ok) return puts("Yes"),0;
}
//��һ���㵥��
ms(bo,0);
bo[1] = true;
int fi = -1;
rep1(i,2,n)
if (!bo[i]){
bo[i] = true;
fi = i;
rep1(j,i+1,n)
if (!bo[j]){
if (ok(1,i,j)) break;
bo[j] = true;
rep1(k,j+1,n)
if (!bo[k] && ok(i,j,k)){
bo[k] = true;
}
break;
}
break;
}
if (fi==-1)
puts("No");
else{
rep1(i,1,n)
if (!bo[i]){
return puts("No"),0;
}
puts("Yes");
}
return 0;
}