传送门 Luogu P3033 [USACO11NOV]牛的障碍Cow Steeplechase
题目描述
Farmer John has a brilliant idea for the next great spectator sport: Cow Steeplechase! As everyone knows, regular steeplechase involves a group of horses that race around a course filled with obstacles they must jump over. FJ figures the same contest should work with highly-trained cows, as long as the obstacles are made short enough.
In order to design his course, FJ makes a diagram of all the N (1 <= N <= 250) possible obstacles he could potentially build. Each one is represented by a line segment in the 2D plane that is parallel to the horizontal or vertical axis. Obstacle i has distinct endpoints (X1_i, Y1_i) and (X2_i, Y2_i) (1 <= X1_i, Y1_i, X2_i, Y2_i <= 1,000,000,000). An example is as follows:
FJ would like to build as many of these obstacles as possible, subject to the constraint that no two of them intersect. Starting with the diagram above, FJ can build 7 obstacles:
Two segments are said to intersect if they share any point in common, even an endpoint of one or both of the segments. FJ is certain that no two horizontal segments in the original input diagram will intersect, and that similarly no two vertical segments in the input diagram will intersect.
Please help FJ determine the maximum number of obstacles he can build.
给出N平行于坐标轴的线段，要你选出尽量多的线段使得这些线段两两没有交点（顶点也算），横的与横的，竖的与竖的线段之间保证没有交点，输出最多能选出多少条线段。
输入输出格式
输入格式
Line 1: A single integer: N.
Lines 2..N+1: Line i+1 contains four space-separated integers representing an obstacle: X1_i, Y1_i, X2_i, and Y2_i.
输出格式Line 1: The maximum number of non-crossing segments FJ can choose.
输入输出样例
输入样例#1
3
4 5 10 5
6 2 6 12
8 3 8 5
输出样例#1
2
说明
There are three potential obstacles. The first is a horizontal segment connecting (4, 5) to (10, 5); the second and third are vertical segments connecting (6, 2) to (6, 12) and (8, 3) to (8, 5).
The optimal solution is to choose both vertical segments.
解题思路
根据题意，我们得出如果有两条线段之间有交点，那么这两条线段就只能选择其中一条，所以我们把有交点的线段之间连边，跑一遍匈牙利，找出最大匹配数，然后用n一减就好了
下面是全洛谷跑的最快内存最小最蒟蒻的代码
C++代码
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#define rr register
#define forvec(T,a,it) for(vector<T>::iterator it=a.begin();it!=a.end();++it)
using namespace std;
const int maxn=255;
int n,sum=0;
struct node{
int pos,far,near;
}heng[maxn],shu[maxn];
int c_heng,c_shu;
int link[maxn],vis[maxn];
inline int read(){
int chtholly=0,willem=1;char c=getchar();
while(c<'0' || c>'9'){if(c=='-')willem=-1;c=getchar();}
while(c<='9' && c>='0'){chtholly=chtholly*10+(int)(c-'0');c=getchar();}
return chtholly*willem;
}
inline bool you_jiao_dian(int i,int j){
if(heng[i].far>=shu[j].pos && heng[i].near<=shu[j].pos && shu[j].far>=heng[i].pos && shu[j].near<=heng[i].pos)
return 1;
return 0;
}
inline bool dfs(int x,vector<int>*a){
forvec(int,a[x],it){
int v=*it;
if(!vis[v]){
vis[v]=1;
if(link[v]==-1 || dfs(link[v],a)){
link[v]=x;return 1;
}
}
}
return 0;
}
inline void max_match(vector<int>*a){
memset(link,0xff,sizeof(link));
for(rr int i=1;i<=c_heng;++i){
memset(vis,0,sizeof(vis));
if(dfs(i,a))sum++;
}
}
int main(){
n=read();
vector<int>a[n+1];
for(rr int i=1;i<=n;++i){
int x1=read(),y1=read(),x2=read(),y2=read();
if(x1==x2){
shu[++c_shu].pos=x1;
shu[c_shu].far=max(y1,y2);
shu[c_shu].near=min(y1,y2);
} else if(y1==y2){
heng[++c_heng].pos=y1;
heng[c_heng].far=max(x1,x2);
heng[c_heng].near=min(x1,x2);
}
}
for(rr int i=1;i<=c_heng;++i)
for(rr int j=1;j<=c_shu;++j)
if(you_jiao_dian(i,j))
a[i].push_back(j);
max_match(a);
printf("%d\n",n-sum);
return 0;
}
