Problem I. Magic Potion There are n heroes and m monsters living in an island. The monsters became very vicious these days, so the heroes decided to diminish the monsters in the island. However, the i-th hero can only kill one monster belonging to th

Magic Potion Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 488    Accepted Submission(s): 287 Problem Description In a distant magic world, there is a powerful magician aswmtjdsj. One day,aswm

题目链接:http://codeforces.com/gym/101981/attachments There are n heroes and m monsters living in an island. The monsters became very vicious these days,so the heroes decided to diminish the monsters in the island. However, the i-th hero can only kill on

Magic Potion http://codeforces.com/gym/101981/attachments/download/7891/20182019-acmicpc-asia-nanjing-regional-contest-en.pdf 从源点到英雄分别拉容量为1和2的边,跑两遍网络流,判断两次的大小和k的大小 #include<iostream> #include<cstring> #include<string> #include<cmath&g

Bit Magic Problem Description Yesterday, my teacher taught me about bit operators: and (&), or (|), xor (^). I generated a number table a[N], and wrote a program to calculate the matrix table b[N][N] using three kinds of bit operator. I thought my ac

原题链接 2018南京的铜牌题,听说学长他们上来就A了,我这个图论选手也就上手做了做,结果一言难尽...... 发此篇博客希望自己能牢记自己的菜... 本题大意:有n个heros和m个monsters大战,每个heros只能杀一个monsters且他们只能杀固定编号的monsters,现在有药水k瓶,如果英雄喝了药水那么它又可以多杀一个怪兽,每个英雄最多能喝一瓶药水.现在给你n, m, k,还有n个英雄各自能击杀怪兽的编号,问这些英雄最多能杀多少怪兽. 题解太水,大佬请绕道... 这里我们将英雄

意甲冠军: a[i] ^ x = f[i] ( i = 1...8 ) 和 ( a[1] + a[2] + ... + a[8] ) ^ x = f[9] 如今f为已知  求x 思路: 从低位到高位确定x值  做法见凝视 代码: #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int f[20],t,ans; int main() { int i,k,j; scanf

题面 题意:n个英雄,m个怪兽,第i个英雄可以打第i个集合里的一个怪兽,一个怪兽可以在多个集合里,有k瓶药水,每个英雄最多喝一次,可以多打一只怪兽,求最多打多少只 n,m,k<=500 题解:显然的最大流裸题,多加一个药水点,药酱入度k,然后再连向英雄 队友抄的模板所以不是我的那个板子 #include<bits/stdc++.h> using namespace std; struct Edge { int from,to,cap,flow; Edge(int u,int v,int

题解:最大流板题:增加两个源点,一个汇点.第一个源点到第二个源点连边,权为K,然后第一个源点再连其他点(英雄点)边权各为1,然后英雄和怪物之间按照所给连边(边权为1). 每个怪物连终点,边权为1: 参考代码: #include<bits/stdc++.h> using namespace std; #define INF 0x3f3f3f3f ; int n,m,k,s,t,u,v,w,num,num1; struct Edge { int from, to, cap, flow; }; ve

http://codeforces.com/gym/101981/attachments 题意:有n个英雄,m个敌人,k瓶药剂,给出每个英雄可以消灭的敌人的编号.每个英雄只能消灭一个敌人,但每个英雄只能消灭一个敌人.现在有药剂,英雄喝了之后可以多消灭一个敌人,但每个英雄只能喝一瓶,问最多能消灭多少个敌人. 下午在实验室队内自己开训练,和JC大佬那队一起开的,当时JC大佬他们队开的J题,没有看I题,当我们队AC之后JC大佬才看了I题,听到他们说,这题就差直接把网络流三个字写在题目里了.确实非常明显

题意: 若干个勇士,每个勇士只能杀特定的怪物.每个勇士只能杀1个怪,但是有一些药,喝了药之后能再杀一个,每个勇士只能喝一瓶药.问你最多杀多少怪. 题解: 按照如下建图套网络流板即可. 网上有题解说套DinicT了,我们队套kuangbin的dinic过了,不得不说kuangbin的板质量真的好. #include<bits/stdc++.h> using namespace std; #define ll long long #define MAXN 1500 //最大点数 #define M

题意: n个英雄,m个怪兽,第i个英雄可以打第i个集合里的怪兽,一个怪兽可以在多个集合里 有k瓶药水,每个英雄最多喝一次,可以多打一只怪兽,求最多打多少只 n,m,k<=500 思路: 最大流,建图方式: 代码: #include<iostream> #include<cstdio> #include<algorithm> #include<cmath> #include<cstring> #include<string> //

题意:类似二分图匹配给的题目,不过这次在这里给出了k,表示没人可以再多一次匹配机会,这次匹配不能用上一次被匹配的对象 分析:不能用匈牙利做俩次匹配,因为俩次的最大匹配并不等价于总和的匹配,事实证明,你用俩次匹配后会被卡在17个样例   既然二分图不能用匈牙利,那么只能考虑用网络流,这里讲到对于k的处理,这里分配的k次得分配完后保证每个英雄至多消灭2个怪兽,所以只要在起点在建一个顶点,连容量为k的边,再在该顶点连线               向英雄容量为1的边 #include<bits/std

题目: 题意:n个士兵打m个怪兽,每个士兵只能打一个,但是如果有魔法药水就可多打一个问最多能打几个. 题解:如果没有魔法药就是一道裸二分图,因为现在有魔法要我们可以这样建图: 多建一个i+n的节点存放内容与i节点一样,怪兽用2*n+p表示:然后用匈牙利算法跑一边1-2*n和1-n比较一下ans2+k与ans1的大小即可. 1 //#include<bits/stdc++.h> 2 #include<time.h> 3 #include <set> 4 #include

Thanks a lot for helping Harry Potter in finding the Sorcerer's Stone of Immortality in October. Did we not tell you that it was just an online game ? uhhh! now here is the real onsite task for Harry. You are given a magrid S ( a magic grid ) having

Magic Grid Time Limit:336MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Description Thanks a lot for helping Harry Potter in finding the Sorcerer's Stone of Immortality in October. Did we not tell you that it was just an online game ? uhhh!

22款制作精美的 iOS 应用程序图标设计作品,遵循图形设计的现代潮流,所有图标都非常了不起,给人惊喜.通过学习这些移动应用程序图标,设计人员可以提高他们的创作,使移动用户界面看起来更有趣和吸引人. 您可能感兴趣的相关文章 45款唯美的苹果 iOS 应用程序图标设计 40款 iPhone 和 iPad 应用程序图标设计 设计前沿:30款超级精美的iOS图标欣赏 10大优秀的移动Web应用程序开发框架 40款 iPhone 和 iPad 应用程序图标设计 GiftBox iOS Icon Magi

https://codeforces.com/gym/101981 Problem A. Adrien and Austin 贪心,注意细节 f[x]=1:先手必赢. f[x]: 分成两部分(或一部分),长度分别为a和b,只要存在f[a] xor f[b]=0,则f[x]=1. #include <bits/stdc++.h> using namespace std; #define ll long long #define minv 1e-6 #define inf 1e9 #define

Thanks a lot for helping Harry Potter in finding the Sorcerer's Stone of Immortality in October. Did we not tell you that it was just an online game ? uhhh! now here is the real onsite task for Harry. You are given a magrid S ( a magic grid ) having

A Thanks a lot for helping Harry Potter in finding the Sorcerer's Stone of Immortality in October. Did we not tell you that it was just an online game ? uhhh! now here is the real onsite task for Harry. You are given a magrid S ( a magic grid ) havin