Alice and Bob
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3046 Accepted Submission(s): 995
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
题目大意：
这道题就是说，给你2组含有n个不同大小的矩形块，要去A的矩形块能够完全覆盖B的矩形块，且每个矩形块只能用一次.
求最多能有多少个矩形被覆盖.
解题思路：
一开始少看了一个0，想用O(n2)试试，看看数据水不水，结果，果断的T掉了。
其实呢，这道题就是一个有关multiset的应用，我们只要对于a和b两个数组按照w的从小到大排序后，依次在b[]中小于a[].w的
加入到multiset中去，因为加入的条件已经满足了w的大小关系，现在我们就用multiset中的成员函数upper_bound去寻找到第
一个比a[i].h大的下标，然后把这个下标以前的全部剪掉，就是我们所有要找到的满足条件的方案数目了。
切记，每次使用完multiset后，都要clear下，要不然会对后面的计算产生不必要的结果.
代码：
1 # include<cstdio>
2 # include<iostream>
3 # include<set>
4 # include<algorithm>
5
6 using namespace std;
7
8 # define MAX 100000+4
9
10 multiset<int>ms;
11
12 struct node
13 {
14 int w,h;
15 }a[MAX],b[MAX];
16
17 int cmp1 ( const struct node & xx, const struct node & yy )
18 {
19 if ( xx.w==yy.w )
20 return xx.h < yy.h;
21 return xx.w < yy.w;
22 }
23
24
25 int main(void)
26 {
27 int t;scanf("%d",&t);
28 while ( t-- )
29 {
30 int n;scanf("%d",&n);
31 for ( int i = 0;i < n;i++ )
32 {
33 scanf("%d%d",&a[i].w,&a[i].h);
34 }
35 for ( int i = 0;i < n;i++ )
36 {
37 scanf("%d%d",&b[i].w,&b[i].h);
38 }
39 sort(a,a+n,cmp1);
40 sort(b,b+n,cmp1);
41 int ans = 0;
42 multiset<int>::iterator it;
43 for ( int i = 0, j = 0;i < n;i++ )
44 {
45 while ( j < n&&b[j].w<=a[i].w )
46 {
47 ms.insert(b[j].h);
48 j++;
49 }
50 it = ms.upper_bound(a[i].h);
51 if ( it!=ms.begin() )
52 {
53 it--;
54 ms.erase(it);
55 ans++;
56 }
57
58 }
59 ms.clear();
60 printf("%d\n",ans);
61 }
62
63 return 0;
64 }