UVA 10763 - Foreign Exchange(二分查找)
来源:互联网 发布:淘宝店铺怎么优化标题 编辑:程序博客网 时间:2024/05/02 01:51
Description
Your non-profit organization (iCORE - international Confederation of Revolver Enthusiasts) coordinates a very successful foreign student exchange program. Over the last few years, demand has sky-rocketed and now you need assistance with your task.
The program your organization runs works as follows: All candidates are asked for their original location and the location they would like to go to. The program works out only if every student has a suitable exchange partner. In other words, if a student wants to go from A to B, there must be another student who wants to go from B to A. This was an easy task when there were only about 50 candidates, however now there are up to 500000 candidates!
Input
The input file contains multiple cases. Each test case will consist of a line containing n - the number of candidates (1≤n≤500000), followed by n lines representing the exchange information for each candidate. Each of these lines will contain 2 integers, separated by a single space, representing the candidate's original location and the candidate's target location respectively. Locations will be represented by nonnegative integer numbers. You may assume that no candidate will have his or her original location being the same as his or her target location as this would fall into the domestic exchange program. The input is terminated by a case where n = 0; this case should not be processed.
Output
For each test case, print "YES" on a single line if there is a way for the exchange program to work out, otherwise print "NO".
Sample Input
10
1 2
2 1
3 4
4 3
100 200
200 100
57 2
2 57
1 2
2 1
10
1 2
3 4
5 6
7 8
9 10
11 12
13 14
15 16
17 18
19 20
0
Sample Output
YES
NO
===================================
关于题意的理解需要注意1 2和2 1是YES,1 2、2 3、3 1是NO
思路就是 快排+二分查找
#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;bool mark[511111];struct student{ int a,b;}s[511111];bool cmp(student x,student y){ if(x.b==y.b) return x.a<y.a; return x.b<y.b;}int search(int key1,int key2,student vis[],int n){ int l=0,r=n-1,m,p=0; while (l<=r) { m=(l+r)/2; if(vis[m].b==key1&&vis[m].a==key2) { p=m;//找到第一个符合相等条件的位置 r=m-1; } else if(vis[m].b>key1||(vis[m].b==key1&&vis[m].a>key2)) r=m-1; else l=m+1; } if(p) { if(!mark[p]) return p; else { while(mark[p]) p++; if(vis[p].b==key1&&vis[p].a==key2) return p; } } return 0;//s[0]肯定会先找,所以这里返回0没有什么问题}int main(){ int n; while(~scanf("%d",&n)) { if(n==0) break; memset(s,0,sizeof(s)); memset(mark,0,sizeof(mark)); int ok=1; for(int i=0;i<n;i++) scanf("%d%d",&s[i].a,&s[i].b); sort(s,s+n,cmp); /*for(int i=0;i<n;i++) { cout<<s[i].a<<"---"<<s[i].b<<endl; }*/ for(int i=0;i<n;i++) { if(!mark[i]) { int p=search(s[i].a,s[i].b,s,n); //cout<<p<<endl; if(p) { mark[i]=1; mark[p]=1; } else { ok=0; break; } } } if(ok) printf("YES\n"); else printf("NO\n"); } return 0;}
- UVA 10763 - Foreign Exchange(二分查找)
- uva 10763 Foreign Exchange(排序+二分查找)
- UVa 10763 - Foreign Exchange
- UVa 10763 - Foreign Exchange
- uva 10763 - Foreign Exchange
- uva 10763 - Foreign Exchange
- Foreign Exchange uva 10763
- UVa 10763 - Foreign Exchange
- Uva-10763-Foreign Exchange
- UVa 10763 - Foreign Exchange
- UVa 10763 - Foreign Exchange
- UVa 10763: Foreign Exchange
- UVA 10763 Foreign Exchange
- UVa 10763 - Foreign Exchange
- uva 10763 - Foreign Exchange
- uva 10763 Foreign Exchange
- uva 10763 Foreign Exchange
- uva 10763 Foreign Exchange
- 关于java和c++中布尔量的比较
- hdu 4336 Card Collector
- hdu 1236
- #include<queue> 【c++】
- 设置和破解Syskey双重加密电脑开机
- UVA 10763 - Foreign Exchange(二分查找)
- 进程与线程及通信总结
- Navicat windowns下远程连接linux端的mysql
- ddddddddddddddddddddddddd
- 堆排序
- python GIL (Global Interpreter Lock)
- 试试看
- request.getParameterValues与request.getParameter的区别
- c语言嵌入式系统修炼之道