题意 : 乱七八糟说了一大堆,实际上就是问你从一个序列到另个序列最少经过多少步的变化,每一次变化只能取序列的任意一个元素去和首元素互换 分析 : 由于只能和第一个元素去互换这种操作,所以没啥最优的特别方法,只要乖乖模拟即可,如果第一个元素不在正确位置则将它和正确位置的元素交换使其回到正确位置,如果第一个元素的位置就是正确的,那么就往后找不在正确位置的元素将它互换到第一个去,然后再对第一个元素进行判断即可,直到序列变成最终满足条件的序列........ #include<bits/stdc++.h…

In this problem, we will define a graph called star graph, and the question is to find the minimum distance between two given nodes in the star graph. Given an integer nn, an n-dimensionaln−dimensionalstar graph, also referred to as S_{n}S​n​​, is an…

In this problem, we will define a graph called star graph, and the question is to find the minimum distance between two given nodes in the star graph. Given an integer nnn, an n−dimensionaln-dimensionaln−dimensional star graph, also referred to as Sn…

2017-09-25 19:58:04 writer:pprp 题意看上去很难很难,但是耐心看看还是能看懂的,给你n位数字 你可以交换第一位和之后的某一位,问你采用最少的步数可以交换成目标 有五组数据 用BFS进行搜索,并进行剪枝,已经搜索过的点不再搜索 #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <string> #…

2000年台湾大专题...英语阅读输入输出专场..我只能说很强势.. M. Frequent Subsets Problem The frequent subset problem is defined as follows. Suppose UU={1, 2,\ldots…,N} is the universe, and S_{1}S​1​​, S_{2}S​2​​,\ldots…,S_{M}S​M​​ are MM sets over UU. Given a positive constan…

题目连接 : https://nanti.jisuanke.com/t/A1256 Life is a journey, and the road we travel has twists and turns, which sometimes lead us to unexpected places and unexpected people. Now our journey of Dalian ends. To be carefully considered are the following…

题目链接 裸的结论题.百度 Ramsey定理.刚学过之后以为在哪也不会用到23333333333,没想到今天网络赛居然出了.顺利在题面更改前A掉~~~(我觉得要不是我开机慢+编译慢+中间暂时死机,我还能再早几分钟过掉它 #include<bits/stdc++.h> using namespace std; ][]; int n; void solve() { ; i<=n; i++) ; j<=n; j++) ; k<=n; k++) { if(g[i][j]==g[i][…

In this problem, we will define a graph called star graph, and the question is to find the minimum distance between two given nodes in the star graph. Given an integer nnn, an n−dimensionaln-dimensionaln−dimensional star graph, also referred to as Sn…

#1636 : Pangu and Stones 时间限制:1000ms 单点时限:1000ms 内存限制:256MB 描述 In Chinese mythology, Pangu is the first living being and the creator of the sky and the earth. He woke up from an egg and split the egg into two parts: the sky and the earth. At the begi…

https://hihocoder.com/problemset/problem/1586 线段树操作,原来题并不难..... 当时忽略了一个重要问题,就是ax*ay要最小时,x.y可以相等,那就简单了!结果显然是模最小的数的平方或者是个负数. 要求乘积最小只要求区间内最大值.最小值和绝对值小的数,再判断min,max正负就行了. 下面的代码相当于建了三个树分别存Min,Max,Mid(abs(Min)) #include<iostream> #include<cstdio> #i…