hdu4268 Alice and Bob(二维贪心的固定思想 STL的multiset的二分查找函数实现)
来源:互联网 发布:相声世界知乎 编辑:程序博客网 时间:2024/06/05 08:42
Link:http://acm.hdu.edu.cn/showproblem.php?pid=4268
Alice and Bob
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3933 Accepted Submission(s): 1230
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.
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.
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
221 23 42 34 532 35 76 84 12 53 4
Sample Output
12
Source
2012 ACM/ICPC Asia Regional Changchun Online
编程思想:二维的贪心思想:先找出满足第一维的所有元素,放到集合里面,再找出最接近第二维的元素,然后从集合中删除该元素。
AC code:
#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>#include<cmath>#include<queue>#include<map>#include<stack>#include<vector>#include<set>#define LL long long#define MAXN 1000010using namespace std;const int N=20;struct node{int h;int w;bool operator < (const node &b)const{if(h!=b.h)return h<b.h;elsereturn w<b.w;}}card1[MAXN],card2[MAXN];int ans;multiset<int>sett;multiset<int>::iterator it;int main(){//freopen("D:\\in.txt","r",stdin);int T,i,j,jj,n;scanf("%d",&T);while(T--){sett.clear();scanf("%d",&n);for(i=1;i<=n;i++){scanf("%d%d",&card1[i].h,&card1[i].w);}for(i=1;i<=n;i++){scanf("%d%d",&card2[i].h,&card2[i].w);}sort(card1+1,card1+n+1);sort(card2+1,card2+n+1);j=1;ans=0;for(i=1;i<=n;i++){while(j<=n&&card1[i].h>=card2[j].h)//找出满足第一维的所有元素 {sett.insert(card2[j].w);//放入集合里面 j++;}if(sett.empty())continue;it=sett.lower_bound(card1[i].w);//lower_bound(val): 返回容器中第一个值【大于或等于】val的元素的iterator位置 if(*it==card1[i].w)//查找的第二维元素满足且最接近(贪心思想:“最接近”在这里的意思是尽可能大的意思) {ans++;sett.erase(it);}else//查找的第二维元素不满足,往前找元素 {if(it==sett.begin()) continue;else{sett.erase(--it);//找第二维满足的且尽可能大的 ans++;}}}printf("%d\n",ans);} return 0;}
0 0
- hdu4268 Alice and Bob(二维贪心的固定思想 STL的multiset的二分查找函数实现)
- [STL][multiset] hdu4268 Alice and Bob
- HDU 4268 Alice and Bob(贪心+Multiset的应用)
- hdu4268 Alice and Bob,multiset
- hdu4268 Alice and Bob(贪心)
- ACM 贪心 STL multiset 二分查找 hdu4268
- HDU4268 Alice and Bob 【贪心】
- 贪心,STL,multiset(Alice and Bob,HDU 4268)
- hdu4268 排序贪心Alice and Bob
- HDU4268 Alice and Bob【贪心+SET】
- hdu4268 Alice and Bob
- HDU4268 Alice and Bob
- hdu4268 Alice and Bob
- hdu4268 Alice and Bob
- hdu 4268 Alice and Bob(贪心+multiset+二分)
- Alice And Bob Hdu--复杂的贪心
- HDOJ 题目4268 Alice and Bob(multiset,贪心)
- hdoj4268 Alice and Bob (multiset +贪心)
- hihoCoder 1228 Mission Impossible 6
- Decode ways
- ANDROID之路-Android_Intent setFlags | 或运算符的含义
- Delegate理解
- 三分钟学会JUnit单元测试
- hdu4268 Alice and Bob(二维贪心的固定思想 STL的multiset的二分查找函数实现)
- TCP坚持定时器
- Harris corner OpenCV源码查询之路
- 关于Java中数值计算的一些误区
- PHP 生成接口数据格式
- Hadoop 问题汇总
- python-生成器
- Adapter
- Oracle数据库JDBC连接