HDU4268 Alice and Bob【贪心+SET】

Alice and Bob

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4719    Accepted Submission(s): 1480

Problem Description

Alice and Bob's game never ends. Today, they introduce a new game. In this game, both of them have N different rectangular cards respectively. Alice wants to use his cards to cover Bob's. The card A can cover the card B if the height of A is not smaller than B and the width of A is not smaller than B. As the best programmer, you are asked to compute the maximal number of Bob's cards that Alice can cover.
Please pay attention that each card can be used only once and the cards cannot be rotated.

Input

The first line of the input is a number T (T <= 40) which means the number of test cases.  
For each case, the first line is a number N which means the number of cards that Alice and Bob have respectively. Each of the following N (N <= 100,000) lines contains two integers h (h <= 1,000,000,000) and w (w <= 1,000,000,000) which means the height and width of Alice's card, then the following N lines means that of Bob's.

Output

For each test case, output an answer using one line which contains just one number.

Sample Input

2 2 1 2 3 4 2 3 4 5 3 2 3 5 7 6 8 4 1 2 5 3 4

Sample Output

1 2

Source

2012 ACM/ICPC Asia Regional Changchun Online

问题链接：HDU4268 Alice and Bob

问题简述：Alice和Bob各有n个矩形，Alice的矩形覆盖多少Bob的矩形。矩形是由高度h和宽度w给出的。

问题分析：

这个问题被归类为贪心，其实就是个排序匹配问题。数一下Alice的矩形覆盖多少Bob的矩形，要尽可能地多。

这需要一个一个去试探，每次尽量匹配尽可能大并且最为接近的矩形。

矩形按照高度和宽度分别排序。

先匹配高度，再匹配宽度。

程序说明：

矩形用pair类型表示，对于排序来说是方便的，代码比较简洁。

集合类multiset不同于set，允许元素重复。同时需要注意的是，集合中的元素已经从小到大排过序。

函数（方法）upper_bound的用法也需要事先掌握。

题记：（略）

参考链接：（略）

AC的C++语言程序如下：

/* HDU4268 Alice and Bob */

#include 
#include 
#include 
#include 

using namespace std;

const int N = 1e5;

typedef pair rect;
rect a[N], b[N];

int main()
{
    int t, n;

    scanf("%d", &t);
    while(t--) {
        multiset s;

        scanf("%d", &n);
        for(int i=0; i= b[j].first) {
                s.insert(b[j].second);
                j++;
            }

            set::iterator iter = s.upper_bound(a[i].second);
            if(s.size() > 0 && iter != s.begin())
                iter--;
            if(s.size() > 0 && *iter <= a[i].second) {
                ans++;
                s.erase(iter);
            }
        }

        printf("%d\n", ans);
    }

    return 0;
}