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
4 题意：给你n个红球，和n个蓝球。以及每一个球的坐标。，现在定义 如果红球的x和y坐标都比蓝球小，那么红球可以和蓝球匹配。一个球只能匹配一次。问这n对球最大可以组成多少个有效pair 思路：先根据坐标关系建立是否能匹配的关系，然后用二分图最大匹配算法的匈牙利算法跑即可。不会的话可以去学习新算法。细节见代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), ‘\0‘, sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2)ans = ans * a % MOD; a = a * a % MOD; b /= 2;} return ans;}
inline void getInt(int* p);
const int maxn = 510;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int n;
int a[maxn];
int b[maxn];
int c[maxn];
int d[maxn];
int can[maxn][maxn];
int vis[maxn];
int linker[maxn];
bool dfs(int x)
{
repd(i, 1, n)
{
if (can[x][i] && vis[i] == 0)
{
vis[i] = 1;
if (linker[i] == 0 || (dfs(linker[i])))// 没使用或者去寻找新的增广路
{
linker[i] = x;
return 1;
}
}
}
return 0;
}
int main()
{
//freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);
//freopen("D:\\common_text\\code_stream\\out.txt","w",stdout);
gbtb;
cin >> n;
repd(i, 1, n)
{
cin >> a[i] >> b[i];
}
repd(i, 1, n)
{
cin >> c[i] >> d[i];
}
repd(i, 1, n)
{
repd(j, 1, n)
{
if (a[i] < c[j] && b[i] < d[j])
{
can[i][j] = 1;
}
}
}
int ans = 0;
repd(i, 1, n)
{
memset(vis, 0, sizeof(vis));
if (dfs(i))
{
ans++;
}
}
cout << ans << endl;
return 0;
}
inline void getInt(int* p) {
char ch;
do {
ch = getchar();
} while (ch == ‘ ‘ || ch == ‘\n‘);
if (ch == ‘-‘) {
*p = -(getchar() - ‘0‘);
while ((ch = getchar()) >= ‘0‘ && ch <= ‘9‘) {
*p = *p * 10 - ch + ‘0‘;
}
}
else {
*p = ch - ‘0‘;
while ((ch = getchar()) >= ‘0‘ && ch <= ‘9‘) {
*p = *p * 10 + ch - ‘0‘;
}
}
}
原文地址：https://www.cnblogs.com/qieqiemin/p/10827472.html