BZOJ 1719--[Usaco2006 Jan] Roping the Field 麥田巨畫
1719: [Usaco2006 Jan] Roping the Field 麥田巨畫
Time Limit: 5 Sec Memory Limit: 64 MBSubmit: 82 Solved: 26
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Description
Farmer John is quite the nature artist: he often constructs large works of art on his farm. Today, FJ wants to construct a giant "field web". FJ‘s field is large convex polygon with fences along the boundary and fence posts at each of the N corners (1 <= N <= 150). To construct his field web, FJ wants to run as many ropes as possible in straight lines between pairs of non-adjacent fence posts such that no two ropes cross. There is one complication: FJ‘s field is not completely usable. Some evil aliens have created a total of G (0 <= G <= 100) grain circles in the field, all of radius R (1 <= R <= 100,000). FJ is afraid to upset the aliens, and therefore doesn‘t want the ropes to pass through, or even touch the very edge of a grain circle. Note that although the centers of all the circles are contained within the field, a wide radius may make it extend outside of the field, and both fences and fence posts may be within a grain circle. Given the locations of the fence posts and the centers of the circles, determine the maximum number of ropes that FJ can use to create his field web. FJ‘s fence posts and the circle centers all have integer coordinates X and Y each of which is in the range 0..1,000,000.
Input
* Line 1: Three space-separated integers: N, G, and R * Lines 2..N+1: Each line contains two space-separated integers that are the X,Y position of a fence post on the boundary of FJ‘s field. * Lines N+2..N+G+1: Each line contains two space-separated integers that are the X,Y position of a circle‘s center inside FJ‘s field.
Output
* Line 1: A single integer that is the largest number of ropes FJ can use for his artistic creation.
最多的線索數．Sample Input
5 3 16 10
10 7
9 1
2 0
0 3
2 2
5 6
8 3
INPUT DETAILS:
A pentagonal field, in which all possible ropes are blocked by three
grain circles, except for the rope between fenceposts 2 and 4.
Sample Output
1HINT
除了籬笆樁2和4之間可以連接繩索，其余均會經過怪圈
題目鏈接：
http://www.lydsy.com/JudgeOnline/problem.php?id=1719
Solution
首先看到題目應該是幾何題無誤（假裝很有道理
看到n<=150感覺似乎暴力也能過。。想想邊數最多也只有22500條。。。
於是這時候應該馬上想到先預處理每條邊是否可以連。。。
直接算圓和線段的交點？感覺應該可以但是似乎不怎麽好寫。。。
考慮題意。。只要線段有部分含於圓內就不能連。。而這個“部分”可以直接認為是線段上與圓心最近的點。。
於是這個預處理就轉化成了求點到線段的最小距離。。這個套套公式就好。。感覺三分也可以但是tle了（可能是我寫炸了。。
預處理完之後，考慮怎麽求答案。。。
要求線段之間不可以相交。。。。
說白了就是不能有1->4 , 2->5這樣的線段同時存在。。考慮DP。。那肯定只能區間DP了。。
n<=150的話O(n^3)還是很輕松的吧
代碼
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<iostream>
#include<queue>
#include<vector>
#include<map>
#define N 20050
#define ept 1e-6
using namespace std;
int n,m;
double R;
struct P{
double x,y;
}a[200],b[200];
int f[200][200];
bool vis[200][200];
double dis(P u,P v){
return sqrt((u.x-v.x)*(u.x-v.x)+(u.y-v.y)*(u.y-v.y));
}
double DIS(P u,P v,P w) {
double space=0;
double a,b,c;
a=dis(u,v);
b=dis(u,w);
c=dis(v,w);
if(c<=ept||b<=ept) {
space=0;
return space;
}
if(a<=ept){
space=b;
return space;
}
if(c*c>=a*a+b*b){
space=b;
return space;
}
if(b*b>=a*a+c*c) {
space=c;
return space;
}
double p=(a+b+c)/2;
double s=sqrt(p*(p-a)*(p-b)*(p-c));
space=2*s/a;
return space;
}
bool judge(P u,P v,P w){
double d1=DIS(u,v,w);
if(d1>R) return 0;
return 1;
}
bool check(P u,P v){
for(int i=1;i<=m;i++)
if(judge(u,v,b[i])) return 0;
return 1;
}
int main(){
scanf("%d%d%lf",&n,&m,&R);
for(int i=1;i<=n;i++)
scanf("%lf%lf",&a[i].x,&a[i].y);
for(int i=1;i<=m;i++)
scanf("%lf%lf",&b[i].x,&b[i].y);
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
if(i==j) continue;
vis[i][j]=check(a[i],a[j]);
}
}
for(int len=3;len<=n;len++){
for(int i=1;i<=n-len+1;i++){
for(int j=i;j<=i+len-1;j++)
f[i][i+len-1]=max(f[i][i+len-1],f[i][j]+f[j][i+len-1]);
if(vis[i][i+len-1]&&(i!=1||i+len-1!=n))
f[i][i+len-1]++;
}
}
printf("%d\n",f[1][n]);
return 0;
}
This passage is made by Iscream-2001.
BZOJ 1719--[Usaco2006 Jan] Roping the Field 麥田巨畫