# 【BZOJ 1038】[ZJOI2008]瞭望塔

【题目链接】:http://www.lydsy.com/JudgeOnline/problem.php?id=1038

【题意】

【题解】

1.在半平面交的直线的交点处
2.在村子往上的投影处;

【完整代码】

#include <bits/stdc++.h>
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 push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%lld",&x)
#define ref(x) scanf("%lf",&x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

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 = 1000;

struct point
{
double x,y;
}points[N];

struct line
{
point a,b;
double k,c;
void get()
{
k = (a.y-b.y)/(a.x-b.x);
c = a.y-k*a.x;
}
}lines[N],sta[N];

int n,top = 0;
double ans = 1e12;

void in()
{
rei(n);
rep1(i,1,n)
ref(points[i].x);
rep1(i,1,n)
ref(points[i].y);
}

bool cmp1(line a,line b)//把线段按照斜率升序排
{
return a.k < b.k;
}

point getintersec(line a,line b)//求两条直线的交点
{
point t;
t.x = (a.c-b.c)/(b.k-a.k);
t.y = t.x*a.k+a.c;
return t;
}

void Insert(line t)//插入一条新的平面
//因为都会是往上的,所以处理起来会简单一点吧
{
while (top>=2)
{
if (getintersec(sta[top-1],sta[top]).x>getintersec(sta[top],t).x) top--;
else
break;
}
sta[++top] = t;
}

void halfpaneintersec()//搞平面交
{
rep1(i,1,n-1)
Insert(lines[i]);
}

void pre()
{
rep1(i,1,n-1)
lines[i].a = points[i],lines[i].b = points[i+1],lines[i].get();
sort(lines+1,lines+1+(n-1),cmp1);
halfpaneintersec();
}

double maxh(double x)//村子的投影往上的交点的纵坐标
{
double t = 0;
rep1(i,1,top)
{
double y = sta[i].k*x+sta[i].c;
t = max(t,y);
}
return t;
}

double jdy(double x)//平面交的直线的交点的横坐标往下的投影的交点纵坐标
{
rep1(i,2,n)
{
if (points[i].x>=x)
return points[i].y-(points[i].y-points[i-1].y)*(points[i].x-x)/(points[i].x-points[i-1].x);
}
return 0;
}

void get_ans()
{
rep1(i,1,n)
ans = min(ans,maxh(points[i].x)-points[i].y);
rep1(i,1,top-1)
{
point t = getintersec(sta[i],sta[i+1]);
ans = min(ans,t.y-jdy(t.x));
}
}

void o()
{
printf("%.3f\n",ans);
}

int main()
{
//freopen("F:\\rush.txt","r",stdin);
in();
pre();
get_ans();
o();
//printf("\n%.2lf sec \n", (double)clock() / CLOCKS_PER_SEC);
return 0;
}

posted @ 2017-10-04 18:45  AWCXV  阅读(117)  评论(0编辑  收藏  举报