AtCoder Regular Contest 092 C - 2D Plane 2N Points 贪心 匈牙利算法模板

Problem Statement
On a two-dimensional plane, there are N red points and N blue points. The coordinates of the i-th red point are (ai,bi), and the coordinates of the i-th blue point are (ci,di).

A red point and a blue point can form a friendly pair when, the x-coordinate of the red point is smaller than that of the blue point, and the y-coordinate of the red point is also smaller than that of the blue point.

At most how many friendly pairs can you form? Note that a point cannot belong to multiple pairs.

Constraints
All input values are integers.
1≤N≤100
0≤ai,bi,ci,di<2N
a1,a2,…,aN,c1,c2,…,cN are all different.
b1,b2,…,bN,d1,d2,…,dN are all different.
Input
Input is given from Standard Input in the following format:

N
a1 b1
a2 b2
:
aN bN
c1 d1
c2 d2
:
cN dN
Output
Print the maximum number of friendly pairs.

Sample Input 1

3
2 0
3 1
1 3
4 2
0 4
5 5
Sample Output 1

2
For example, you can pair (2,0) and (4,2), then (3,1) and (5,5).

Sample Input 2

3
0 0
1 1
5 2
2 3
3 4
4 5
Sample Output 2
2
For example, you can pair (0,0) and (2,3), then (1,1) and (3,4).

Sample Input 3

2
2 2
3 3
0 0
1 1
Sample Output 3

0
It is possible that no pair can be formed.

Sample Input 4

5
0 0
7 3
2 2
4 8
1 6
8 5
6 9
5 4
9 1
3 7
Sample Output 4

5
Sample Input 5

5
0 0
1 1
5 5
6 6
7 7
2 2
3 3
4 4
8 8
9 9
Sample Output 5
Copy
4
题意：给出n个红色坐标，n个蓝色坐标，如果某一个蓝色坐标大于某一个红色坐标，则构成一对坐标对，每个坐标只可以用一次
，求最多有多少坐标对
思路：
1.贪心：用符合条件最小的q去匹配符合条件最大的p 
类似题目：
校赛 搬书  http://blog.csdn.net/deepseazbw/article/details/78836942
HDU - 1051 Wooden Sticks http://blog.csdn.net/deepseazbw/article/details/78820869
#include <bits/stdc++.h>
using namespace std;
const int N = 205;
typedef long long ll;
struct node
{
int a;
int b;
} p
,q
;
int vis
;
bool cmp(node x,node y)
{
if(x.a!=y.a)
return x.a<y.a;
else
return x.b<y.b;
}

int main()
{
int n;
cin>>n;
for(int i=1; i<=n; i++)
scanf("%d%d",&p[i].a,&p[i].b);
for(int i=1; i<=n; i++)
scanf("%d%d",&q[i].a,&q[i].b);
memset(vis,0,sizeof(vis));
sort(p+1,p+n+1,cmp);
sort(q+1,q+n+1,cmp);
int cnt=0;
//贪心用符合条件最小的q去装符合条件最大的p
for(int j=1; j<=n; j++)
{
int mid=-1,flag=0;
for(int i=1; i<=n; i++)
{
if(!vis[i]&&p[i].a<q[j].a&&p[i].b<q[j].b) //先保证存在
{
flag++;
if(flag==1)
mid=i;
else if(p[i].b>p[mid].b)
mid=i;
}
}
if(mid>0)
{
vis[mid]=1;
cnt++;
}
}
printf("%d\n",cnt);
return 0;
}
2.暴力枚举 DFS 超时 列举出所有的情况#include <bits/stdc++.h>
using namespace std;
const int N = 205;
typedef long long ll;
struct node
{
int a;
int b;
} p
,q
;
int vis
;
int sum,n,ans;
bool cmp(node x,node y)
{
if(x.a!=y.a)
return x.a<y.a;
else
return x.b<y.b;
}
void dfs(int qbegin,int ans)
{
if(qbegin>n)
{
sum=max(sum,ans);
return ;
}
for(int j=qbegin; j<=n; j++)
{
for(int i=1; i<=n; i++)
{
if(q[j].a>p[i].a&&q[j].b>p[i].b&&!vis[i])
{
ans++;
vis[i]=1;
dfs(j+1,ans);
vis[i]=0;
ans--;
}
}
}
}
int main()
{

while(cin>>n)
{
for(int i=1; i<=n; i++)
scanf("%d%d",&p[i].a,&p[i].b);
for(int i=1; i<=n; i++)
scanf("%d%d",&q[i].a,&q[i].b);
memset(vis,0,sizeof(vis));
sort(p+1,p+n+1,cmp);
sort(q+1,q+n+1,cmp);
sum=0;
dfs(1,0);
printf("%d\n",sum);
}
return 0;
}
3.匈牙利算法的模板题目
参考博客
http://blog.csdn.net/cillyb/article/details/55511666
http://blog.csdn.net/qq_35935435/article/details/54906008
#include <bits/stdc++.h>
#include <string>
using namespace std;

typedef long long ll;
const int maxn = 200005;
const ll mod = 1e9+7;
const int MAX_int = 1e9;
const double eps = 1e-9;
int vis[105];
int relation[105][105];
int match[105];
int n;
struct node
{
int a,b;
} p[105],q[105];

bool find(int x)
{
for (int j=1; j<=n; j++)
{
if (relation[x][j]==1 && vis[j]==0)
{
vis[j]=1;
if (match[j]==0 || find(match[j]))
{
//名花无主或者能腾出个位置来，这里使用递归
match[j]=x;
return true;
}
}
}
return false;
}

int main()
{
memset(match,0,sizeof(match));
memset(relation,0,sizeof(relation));
scanf("%d",&n);
for(int i=1; i<=n; i++)
scanf("%d%d",&p[i].a,&p[i].b);
for(int i=1; i<=n; i++)
scanf("%d%d",&q[i].a,&q[i].b);
for(int i=1; i<=n; i++)
for(int j=1; j<=n; j++)
if(p[i].a <q[j].a && p[i].b <q[j].b)
relation[i][j]=1;

int cnt=0;
for (int i=1; i<=n; i++)
{
memset(vis,0,sizeof(vis)); //这个在每一步中清空
if (find(i))
cnt++;
}
printf("%d\n",cnt);
return 0;
}