题意:从s到m的最短时间.(“o"不能走,‘#’走一个花两个单位时间,‘.'走一个花一个单位时间) 思路:广搜和优先队列. #include <stdio.h> #include <string.h> #include <iostream> #include <queue> #define MAX 30 using namespace std; struct pos { int x; int y; int step; }; bool operato

传送门 Ice Cream Parlor Authored by dheeraj on Mar 21 2013 Problem Statement Sunny and Johnny together have M dollars they want to spend on ice cream. The parlor offers N flavors, and they want to choose two flavors so that they end up spending the whol

ShareThis - By Vikas Verma Bluetooth low energy (BLE) is a feature of Bluetooth 4.0 wireless radio technology, aimed at new, principally low-power and low-latency, applications for wireless devices within a short range. As I discussed in my previous

A. Free Ice Cream 题目连接: http://www.codeforces.com/contest/686/problem/A Description After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer. At the start of the day they hav

2017-08-18 21:53:38 writer:pprp 题意如下: Problem D. Ice Cream Tower Input file: Standard Input Output file: Standard Ouptut Time limit: seconds Mr. Panda likes ice cream very much especially the ice cream tower. An ice cream tower consists of K ice crea

/** 题目:2016-2017 ACM-ICPC CHINA-Final Ice Cream Tower 链接:http://codeforces.com/gym/101194 题意:给n个木块,堆一个小塔要k个木块,满足相邻两个木块,上面的木块大小至少是下面的木块的两倍. 问最多可以堆出几个小塔. 思路:二分+贪心. 先二分最终可以堆出的小塔数x,然后确定了数量,就可以这样来贪心,把前x个木块作为所有小塔的顶部木块, 然后从剩下的木块中继续取前x小的,从小到大放在x个小塔的次顶部,然后继续.

E. Sonya and Ice Cream time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Sonya likes ice cream very much. She eats it even during programming competitions. That is why the girl decided that

Sunny and Johnny together have M dollars which they intend to use at the ice cream parlour. Among N flavors available, they have to choose two distinct flavors whose cost equals M. Given a list of cost of N flavors, output the indices of two items wh

传送门 分析 这道题做了好长时间,题意就很难理解. 我们注意到这句话Vertices which have the i-th (1 ≤ i ≤ m) type of ice cream form a connected subgraph 也就是说在最后的图中相同颜色的点构成一个连通图 也就是说如果1和2.3在不同的点共同出现,那么2和3一定不会在某个点共同出现.于是我们可以直接暴力dfs,在每个点对没有颜色的冰激凌贪心染最小的颜色. 需要注意有某种冰激凌没有出现的情况. trick 代码 #in

http://codeforces.com/contest/805/problem/E [题意] 染色数是很好确定,最少染色数是max(si)(最小为1,即使所有的si都为0,这样是单节点树形成的森林需要1种颜色),关键是确定染色方案. 一开始没看出来树有什么用,但其实一句话很关键:Vertices which have the i-th (1 ≤ i ≤ m) type of ice cream form a connected subgraph. 也就是说在原来的图中,含有相同的冰淇淋的点是

题目链接: A. Free Ice Cream //#include <bits/stdc++.h> #include <vector> #include <iostream> #include <queue> #include <cmath> #include <map> #include <cstring> #include <algorithm> #include <cstdio> using

题目: To encourage visitors active movement among the attractions, a circular path with ice cream stands was built in the park some time ago. A discount system common for all stands was also introduced. When a customer buys ice cream at some stand, he

题目: Mr. Panda likes ice cream very much especially the ice cream tower. An ice cream tower consists of K ice cream balls stacking up as a tower. In order to make the tower stable, the lower ice cream ball should be at least twice as large as the ball

Ice cream coloring 题解: 这个题目中最关键的一句话是, 把任意一种类型的冰激凌所在的所有节点拿下来之后,这些节点是一个连通图(树). 所以就不会存在多个set+起来之后是一个新的完全图. 所以只要直接去做就好了. 对于每个节点来说,染色. 代码: #include<bits/stdc++.h> using namespace std; #define Fopen freopen("_in.txt","r",stdin); freope

被一道数位DP折磨得欲仙欲死之后,再做这道题真是如同吃了ice cream一样舒畅啊 #include<bits/stdc++.h> using namespace std; #define rep(i,a,b) for(int i=a;i<=b;++i) #define ms(arr,a) memset(arr,a,sizeof arr) #define debug(x) cout<<"< "#x" = "<<x&l

对那个树进行dfs,在动态维护那个当前的冰激凌集合的时候,显然某种冰激凌仅会进出集合各一次(因为在树上形成连通块). 于是显然可以对当前的冰激凌集合贪心染色.暴力去维护即可.具体实现看代码.map不必要. #include<cstdio> #include<set> #include<vector> #include<map> #include<algorithm> using namespace std; map<int,bool>

题目链接:http://codeforces.com/contest/805/problem/E 题意:你有n个节点,这个n个节点构成一棵树.每个节点拥有有si个类型的ice,同一个节点的ice互相连边构成完全图.对于拥有相同ice的节点u,v,在树上一定相邻.求将ice构成的图染色,相邻点不可同色的最小颜色数以及方案. 题解:就是简单的dfs一遍现将首个节点对应的ice赋予不同的数字,记录一下ice对应的颜色,再dfs到下个树节点,现将 已经染过色的ice存一下然后再遍历一遍没染过色的ice给

题目大意: 对于给出的n个冰激凌球的大小,满足下面的球的大小是上一个的至少2倍,对于给出的k(由k的冰激凌球才能算作一个冰激凌塔),问n个冰激凌球可以最多堆出多少个高度为k的冰激凌塔 题目分析: 对于n个冰激凌球,显然我们得知可以堆出的高度为k的塔的数量在0~[n / k]之间,这里可以通过二分遍历每一种可能,初始时二分边界l==0,r==[n / k],每次取中间值mid=(l+r)/ 2,判断mid高度为k的塔能否堆出,如果可以则尝试mid为更大,否则则尝试mid为更小时,不断二分尝试mid

题目: 儿童排队领冰激凌,给你两个数n,x分别代表接下来有n行与初始的冰激淋数:接下来n行,每行有一个字符('+'or‘-’),还有一个整数d,+d表示新增的冰激 凌数(由搬运工搬运到此),-d表示儿童将要领走的冰激凌数(当剩余的数量有足够这么多时),求没领到冰激凌的儿童. 分析: 由案例可以知道当目前的冰激凌数不足以分发给当前儿童所要的数目d时,该儿童不能领到冰激凌 代码如下: #include <iostream> #include <cstdio> #include <

题目链接: http://codeforces.com/gym/101194/attachments https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=5922 题意: 给出 $N$ 个冰淇淋球,第 $i$ 个冰淇淋球大小为 $B_i$,现在已知冰淇淋球堆叠起来可组成一个冰淇淋. 对于上下相邻的两个冰淇淋球,只有上面的那个大小不

题目链接: http://codeforces.com/gym/101194/attachments https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=5923 题意: 现有 $N$ 支队伍参加比赛,只有一个队伍能获胜.给出每个队伍一个赔率 $A_i:B_i$,你往这个队投 $x$ 元,若该队获胜你可得到 $\frac{A_i+