An interesting question 2 --- sort of thinking, enumerate

 

　　This question is another way to listen and learn to eat school brother brother speak, I feel pretty good ^ _ ^, such as if the younger brother often hear a lot of talk about the title can harvest it, probably next year is expected written, ha ha ~

　　Problem:

　　　　N-plane to the individual, the position (x, y) given tuple, m plane there door, a position (x, y) given. Now the agreement can only enter every door a man, and man can only move left and down (to x-1 y-1 and moving), I ask how many people are up to the door?

 

　　Solution：

　　　　The door and the x value by descending order, the enumeration door. For the current door enumerated i, is greater than the door [i] y into the value of all .x set, the minimum is found greater than or equal door [i] .y, and assign it to the gate i, and then from set excluded, then the door enumeration i + 1.

 

　　code show as below:

　　

#include <the iostream> 
#include <algorithm> 
#include < SET >
 the using  namespace STD;
 / * 
Problem: to the plane of the n-individual, the position (x, y) tuple given plane there door m , from the location (x, y) given. Now the agreement can only enter every door a man, and man can only move left and down (to x-1 y-1 and mobile), will at most 
        how many people the door? 
Solution: The door and the x value by descending order, the enumeration door. For the current door enumerated i, is greater than the door [i] y into the value of all .x set, the minimum is found greater than or equal door [i] .y, and assign it to the gate i, and then from set 
        excluded, then the door enumeration i + 1. 
* / 
Const  int N = 1E3 + . 5 ; 
typedef pair < int , int > Tuple;
 BOOL CMP ( const Tuple A, const Tuple B) {
    return a.first > b.first;
}

multiset<int> s;
multiset<int>::iterator it;

int main()
{
    int n, m; cin >> n >> m;
    Tuple people[N], door[N];
    for (int i = 0; i < n; i++) {
        cin >> people[i].first >> people[i].second;
    }
    sort(people, people + n, cmp);
    for (int i = 0; i < m; i++) {
        cin >> door[i].first >> door[i].second;
    }
    sort(door, door + m, cmp);
    
    int ans = 0;
    int k = 0;
    for (int i = 0; i < m; i++) {
        while (k < n) {
            if (people[k].first >= door[i].first) {
                s.insert(people[k].second);
                k++;
            }
        }
        if (s.count(door[i].second) > 0) {
            ans++;
            s.erase(s.find(door[i].second));
        }
        else {
            it = s.upper_bound(door[i].second);
            if (it != s.end()) {
                ans++;
                s.erase(it);
            }
        }
    }
    std::cout << "Answer = " << ans << endl;
    return 0;
}

/*
2 2
5 3
6 5
3 4
4 2
*/