Codeforces 902 A.Visiting a Friend 贪心
2017-12-20 16:48
225 查看
题意
给出n(100),m(100),然后是n个a,b(0<=a<=b<=m),且a非严格单调递增表示在坐标轴上有n个在ai处的传送门,每个可以传送到[ai,bi]的任意位置,问是否可以仅通过传送门从0到达m.
解法
题面很绕,但是注意到如果位置x可达,那么[0,x]全部可达,所以仅需维护一个最远可达的距离,按顺序遍历传送门即可.代码
/* LittleFall : Hello! */ #include <bits/stdc++.h> using namespace std; int main(void) { int n,m; scanf("%d%d",&n,&m); int most=0; for(int i=0;i<n;i++) { int ta,tb; scanf("%d%d",&ta,&tb); if(ta<=most) most=max(tb,most); } if(most>=m) printf("YES\n"); else printf("NO\n"); return 0; }
代码用时:2分钟
注意
其实没有很能理清这个逻辑,但是确实是正确的.a按大小排序很重要.
相关文章推荐
- Visiting a Friend CodeForces - 902A
- CodeForces 732B Cormen — The Best Friend Of a Man (贪心)
- Codeforces 732B Cormen — The Best Friend Of a Man【贪心】
- CodeForces 732 B.Cormen — The Best Friend Of a Man(贪心)
- Codeforces 733D Kostya the Sculptor 贪心
- Codeforces 2015-2016 ACM-ICPC, NEERC, Southern Subregional Contest B题 (排序贪心)
- CodeForces 479C Exams 贪心
- Codeforces 725D Contest Balloons【贪心+优先队列】
- Codeforces 767E Change-free【贪心+优先队列】
- CodeForces - 803C Maximal GCD(贪心)
- CodeForces 337C Captains Mode(dp+位运算+贪心)
- CodeForces 35 D.Animals(贪心)
- codeforces 3B (贪心)
- Codeforces 584E Anton and Ira【思维+贪心】好题~
- CodeForces CF 360E Levko and Game 贪心+SPFA
- codeforces 352 div 2 C.Recycling Bottles 贪心
- Codeforces 796C Bank Hacking 贪心+规律
- Codeforces_389B_Fox and Cross(贪心)
- CodeForces 731B-Coupons and Discounts(贪心 模拟)
- codeforces 712C C. Memory and De-Evolution(贪心)