Codeforces 845 C Two TVs
2017-08-23 01:11
477 查看
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
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|1 using namespace std; const LL mod = 1e9 + 7; struct node { int st, en; }TV ; 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 <& bf0d lt; "YES" << endl; } else { cout << "NO" << endl; } } return 0; }
相关文章推荐
- Codeforces 845 C Two TVs(贪心之优先队列维护)
- Codeforces-845C:Two TVs(思维)
- CodeForces 845C Two TVs (模拟)
- codeforces 620D D. Professor GukiZ and Two Arrays
- Codeforces 702B - Powers of Two
- 【CodeForces 632B】 Alice, Bob, Two Teams(暴力)
- Codeforces 845 G Shortest Path Problem?
- Codeforces Round #179 (Div. 2) B (codeforces 296b) Yaroslav and Two Strings
- dfs:Codeforces--14D--Two Paths(树的直径)
- codeforces 510.B Fox And Two Dots (DFS好题)
- Codeforces 632B:Alice, Bob, Two Teams(英文题。。。)
- CodeForces - 520B Two Buttons
- CodeForces 632B- Alice, Bob, Two Teams
- CodeForces 616A Comparing Two Long Integers【字符串】
- G - Two Buttons CodeForces - 520B
- codeforces 520 Two Buttons
- CodeForces 520 B.Two Buttons(bfs)
- [DFS] Codeforces 510B:Fox And Two Dot
- Codeforces--14D--Two Paths(树的直径)
- Codeforces 845 A Chess Tourney