Codeforces 845 C Two TVs
来源:互联网 发布:mysql和oracle语法区别 编辑:程序博客网 时间:2024/05/18 08:55
题目地址
题意:你是土豪,你有两台电视机,你有n个想看的节目,在同一台电视机上不能看a~b时间和b~c时间的,就是说不能看一档节目的结束时间等于另一档节目的开始时间的节目。
思路:按照时间进行排序,维护两台电视机现在播放的节目的结束时间就好了,直接模拟求,详细看代码
#include <iostream>#include <cstring>#include <string>#include <queue>#include <vector>#include <map>#include <set>#include <stack>#include <cmath>#include <cstdio>#include <algorithm>#define N 200010#define M 90010#define LL __int64#define inf 0x3f3f3f3f#define lson l,mid,ans<<1#define rson mid+1,r,ans<<1|1#define getMid (l+r)>>1#define movel ans<<1#define mover ans<<1|1using namespace std;const LL mod = 1e9 + 7;struct node { int st, en;}TV[N];bool cmp(node a, node b) { if (a.st == b.st) { return a.en < b.en; } return a.st < b.st;}int main() { cin.sync_with_stdio(false); int n; int tv1, tv2; bool flag; while (cin >> n) { tv1 = -1; tv2 = -1; flag = true; for (int i = 0; i < n; i++) { cin >> TV[i].st >> TV[i].en; } sort(TV, TV + n, cmp); for (int i = 0; i < n&&flag; i++) { if (tv1 < TV[i].st) { tv1 = TV[i].en; } else { if (tv2 < TV[i].st) { tv2 = TV[i].en; } else { flag = false; } } } if (flag) { cout << "YES" << endl; } else { cout << "NO" << endl; } } return 0;}
阅读全文
0 0
- codeforces 845C Two TVs
- Codeforces 845 C Two TVs
- Codeforces 845C Two TVs【思维】水题
- CodeForces 845C Two TVs (模拟)
- Codeforces-845C:Two TVs(思维)
- Codeforces 845 C. Two TVs 思路:简单贪心算法
- Educational Codeforces Round 27 C. Two TVs(模拟)
- cfC. Two TVs
- CodeForces 762C Two strings
- 【Codeforces 837C. Two Seals】
- CodeForces 425C Sereja and Two Sequences
- Educational Codeforces Round 11 C (Two Pointers)
- 【50.00%】【codeforces 602C】The Two Routes
- [codeforces] 762C - Two strings 线段树
- Educational Codeforces Round 17-C Two strings
- Educational Codeforces Round 26 C. Two Seals
- Educational Codeforces Round 26 C. Two Seals
- Educational Codeforces Round 26 C. Two Seals
- python2/3---sort方法与sorted函数的使用
- Android 监听软键盘弹出
- Xor Sum Gym
- Windows Practice_文件搜索器(四)_封装文件扫描器
- Gym
- Codeforces 845 C Two TVs
- java lambda表达式
- Windows Practice_闹钟(一)_简易记事本
- 门面模式
- k-近邻算法
- $n$-Way Tie Gym
- python学习(一)之初探
- 关于Arduino与STM32
- Qt浅谈之十七:飞舞的蝴蝶(GraphicsView框架)