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| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 6198 | Accepted: 1744 |
Description
Input
Output
Sample Input
6 0 0 8 3 1 4 3 2 2 1 7 1 4 1 2 3 3 5 4 6 2 3 9 8 3 3 0 10 2 5 5 20 25 7 -3 30 32 0
Sample Output
Forest 1 Cut these trees: 2 4 5 Extra wood: 3.16 Forest 2 Cut these trees: 2 Extra wood: 15.00
/*
poj 1873 凸包+枚举
给你n棵树,已知树的树的长度以及他们的价值。要砍掉一些树来给剩下的数围一个篱笆
要求剩下的数价值尽可能大,如果价值相同则希望剩下的树尽可能多
因为最多15棵,枚举需要砍掉的树,然后通过凸包判断是否能围成,记录一下即可
hhh-2016-05-07 21:27:41
*/
#include <iostream>
#include <vector>
#include <cstring>
#include <string>
#include <cstdio>
#include <queue>
#include <cmath>
#include <algorithm>
#include <functional>
#include <map>
using namespace std;
#define lson (i<<1)
#define rson ((i<<1)|1)
typedef long long ll;
using namespace std;
const int maxn = 20;
double PI = 3.1415926;
double eps = 1e-8;
int sgn(double x)
{
if(fabs(x) < eps) return 0;
if(x < 0)
return -1;
else
return 1;
}
struct Point
{
double x,y;
Point() {}
Point(double _x,double _y)
{
x = _x,y = _y;
}
Point operator -(const Point &b)const
{
return Point(x-b.x,y-b.y);
}
double operator ^(const Point &b)const
{
return x*b.y-y*b.x;
}
double operator *(const Point &b)const
{
return x*b.x + y*b.y;
}
};
struct Line
{
Point s,t;
Line() {}
Line(Point _s,Point _t)
{
s = _s;
t = _t;
}
pair<int,Point> operator &(const Line&b)const
{
Point res = s;
if( sgn((s-t) ^ (b.s-b.t)) == 0) //通过叉积判断
{
if( sgn((s-b.t) ^ (b.s-b.t)) == 0)
return make_pair(0,res);
else
return make_pair(1,res);
}
double ta = ((s-b.s)^(b.s-b.t))/((s-t)^(b.s-b.t));
res.x += (t.x-s.x)*ta;
res.y += (t.y-s.y)*ta;
return make_pair(2,res);
}
};
Point lis[maxn];
int Stack[maxn],top;
double dist(Point a,Point b)
{
return sqrt((a-b)*(a-b));
}
Point ta[20];
bool cmp(Point a,Point b)
{
double t = (a-ta[0])^(b-ta[0]);
if(sgn(t) == 0)
{
return dist(a,ta[0]) <= dist(b,ta[0]);
}
if(sgn(t) < 0)
return false;
else
return true;
}
int tot;
double Graham(int n)
{
Point p;
if(n == 1 || n == 0)
{
return 0;
}
if(n == 2)
{
return dist(ta[0],ta[1])*2;
}
int k = 0;
p = ta[0];
for(int i = 1; i < n; i++)
{
if(p.y > ta[i].y || (p.y == ta[i].y && p.x > ta[i].x))
p = ta[i],k = i;
}
swap(ta[0],ta[k]);
sort(ta+1,ta+n,cmp);
Stack[0] = 0;
Stack[1] = 1;
top = 2;
for(int i = 2; i < n; i++)
{
while(top > 1 && sgn((ta[Stack[top-1]]-ta[Stack[top-2]])
^ (ta[i]-ta[Stack[top-2]])) <= 0)
top --;
Stack[top++] = i;
}
double len = 0;
for(int i = 0; i < top; i++)
{
if(i == top - 1)
len += dist(ta[Stack[i]],ta[Stack[0]]);
else
len += dist(ta[Stack[i]],ta[Stack[i+1]]);
}
return len;
}
int val[maxn];
double lent[maxn];
int main()
{
//freopen("in.txt","r",stdin);
int n;
int cas = 1;
while(scanf("%d",&n) && n)
{
if(cas != 1)
printf("\n");
for(int i = 0; i < n; i++)
{
scanf("%lf%lf%d%lf",&lis[i].x,&lis[i].y,&val[i],&lent[i]);
}
int ansV = 0x7fffffff,ansN = 0x7fffffff,ansX = 0;
double ansL = 0;
for(int i = 0; i < (1<<n); i++)
{
tot = 0;
double lans = 0;
int vans = 0;
for(int j = 0; j < n; j++)
{
if(i & (1 << j))
{
vans += val[j];
lans += lent[j];
}
else
{
ta[tot++] = lis[j];
}
}
if(vans > ansV)
continue;
double t = Graham(tot);
if(lans >= t)
{
if(vans < ansV || (n-tot < ansN && vans == ansV))
{
ansV = vans;
ansL = lans-t;
ansX = i;
ansN = n-tot;
}
}
}
printf("Forest %d\n",cas++);
printf("Cut these trees:");
for(int i = 0; i < n; i++)
{
if(ansX&(1 << i))
printf(" %d",i+1);
}
printf("\n");
printf("Extra wood: %.2f\n",ansL);
}
return 0;
}
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原文地址：http://www.cnblogs.com/Przz/p/5510370.html
