B. Kefa and Company Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/580/problem/B Description Kefa wants to celebrate his first big salary by going to restaurant. However, he needs company. Kefa has n friends, each friend wi…
->链接在此<- B. Kefa and Company time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Kefa wants to celebrate his first big salary by going to restaurant. However, he needs company. Kefa has n fri…
排序以后枚举尾部.尺取,头部单调,维护一下就好. 排序O(nlogn),枚举O(n) #include<bits/stdc++.h> using namespace std; typedef long long ll; //#define LOCAL ; struct Node { int m,s; void IN(){ scanf("%d%d",&m,&s); } bool operator < (const Node&rh) const {…
题目链接:http://codeforces.com/problemset/problem/180/E 给你n个数,每个数代表一种颜色,给你1到m的m种颜色.最多可以删k个数,问你最长连续相同颜色的序列的长度是多少. 将相同颜色的下标存到对应颜色的容器中,比如ans[a[i]].push_back(i)就是将下标为i颜色为a[i]的数存到ans[a[i]]容器中. 对于每种颜色序列,尺取一下 在差距小于k的情况下取能取到的最大长度. //#pragma comment(linker, "/STA…
E. Kefa and Watch Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/580/problem/E Description One day Kefa the parrot was walking down the street as he was on the way home from the restaurant when he saw something glittering b…
C. Kefa and Park Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/580/problem/C Description Kefa decided to celebrate his first big salary by going to the restaurant. He lives by an unusual park. The park is a rooted tree con…
A. Kefa and First Steps Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/580/problem/A Description Kefa decided to make some money doing business on the Internet for exactly n days. He knows that on the i-th day (1 ≤ i ≤ n) h…
题目链接: 题目 D. Kefa and Dishes time limit per test:2 seconds memory limit per test:256 megabytes 问题描述 When Kefa came to the restaurant and sat at a table, the waiter immediately brought him the menu. There were n dishes. Kefa knows that he needs exactly…
题目链接:http://www.codeforces.com/problemset/problem/580/A题意:求最长连续非降子序列的长度.C++代码: #include <iostream> using namespace std; ; int n, a[maxn], tmp, ans; int main() { cin >> n; ; i < n; i ++) cin >> a[i]; tmp = ans = ; ; i <= n; i++) ])…
http://codeforces.com/contest/580/problem/D 题意: 有个人去餐厅吃饭,现在有n个菜,但是他只需要m个菜,每个菜只吃一份,每份菜都有一个欢乐值.除此之外,还有一些规则,x,y,w代表的是如果x吃完后吃y,那么还能获得额外的w欢乐值.计算所能获得的最大欢乐值. 思路: 看到别人说要用状压dp来做,我才恍然大悟啊,感觉自己对于状压dp实在是太不敏感了. d[i][j]表示在当前i状态时最后一份吃的是j的最大欢乐值. 状态转移什么的请看代码吧. #includ…