Alice and Bob(贪心HDU 4268)

Alice and Bob
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3716 Accepted Submission(s): 1179

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
贪心的策略,先进行排序,(两个元素谁优先都一样),对bo[i].y(Bob)进行查找，找到最接近al[i].y(Alice)的值，删除他。

#include <iostream>#include <set>#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>using namespace std;const int MAX=101000;struct node{    int w;    int h;    bool operator <(const node &a)const//重载小于号    {        if(w<a.w||(w==a.w&&h<a.h))        {            return true;        }        return false;    }}al[MAX],bo[MAX];multiset<int>st;int main(){    int n;    int T;    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        st.clear();        for(int i=0;i<n;i++)        {            scanf("%d %d",&al[i].h,&al[i].w);        }        for(int i=0;i<n;i++)        {            scanf("%d %d",&bo[i].h,&bo[i].w);        }        sort(al,al+n);        sort(bo,bo+n);        int ans=0;        int j=0;        multiset<int>::iterator it;        for(int i=0;i<n;i++)        {            while(j<n&&bo[j].w<=al[i].w)//将al[i]可能覆盖的放在set中            {                st.insert(bo[j].h);                j++;            }            if(st.empty())            {                continue;            }            it=st.lower_bound(al[i].h);//二分查找            if(*it==al[i].h)//删除            {                ans++;                st.erase(it);            }            else            {                if(it!=st.begin())//如果返回第一个比它大的                {                    st.erase(--it);//删除后面的                    ans++;                }            }        }        printf("%d\n",ans);    }    return 0;}