Outlets                                                                                                              Time Limit: 2000/1000 MS (Java/Others)                                                                                               …

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4463 很裸的一道题目,稍微处理一下输入即可 代码: #include<iostream> #include<cstdlib> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> using namespace std; #define maxn 60…

1.LightOJ 1012  Guilty Prince  简单bfs 2.总结:水 题意:迷宫,求有多少位置可去 #include<iostream> #include<cstring> #include<cmath> #include<queue> #include<algorithm> #include<cstdio> #define F(i,a,b) for (int i=a;i<=b;i++) using names…

串的简单处理 Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 39  Solved: 11[Submit][Status][Web Board] Description 串的处理在实际的开发工作中,对字符串的处理是最常见的编程任务.本题目即是要求程序对用户输入的串进行处理.具体规则如下:1.    把每个单词的首字母变为大写.2.    把数字与字母之间用下划线字符(_)分开,使得更清晰3.    把单词中间有多个空格的调整为1个空格. Input…

Problem Description There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B…

Description The cows have run <= N <= ,) farms (numbered ..N); Bessie starts at Farm . She'll traverse some or all of the M (1 <= M <= 10,000) two-way roads whose length does not exceed 1,000,000,000 that connect the farms. Some farms may be m…

Problem Description In China, foreign brand commodities are often much more expensive than abroad. The main reason is that we Chinese people tend to think foreign things are better and we are willing to pay much for them. The typical example is, on t…

分析:就是判断简单的前缀有没有相同,注意下自身是m的倍数,以及vis[0]=true; #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <iostream> #include <algorithm> #include <map> #include <queue> #include <vect…

题意:给定n个点坐标,并且两个点已经连接,但是其他的都没有连接,但是要找出一条最短的路走过所有的点,并且路线最短. 析:这个想仔细想想,就是应该是最小生成树,把所有两点都可以连接的当作边,然后按最小生成树,写就OK了. 代码如下: #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstr…

以(x,y)坐标的形式给出n个点,修建若干条路使得所有点连通(其中有两个给出的特殊点必须相邻),求所有路的总长度的最小值. 因对所修的路的形状没有限制,所以可看成带权无向完全图,边权值为两点间距离.因是稠密图,故用邻接矩阵存储更好(完全图,边数e达到n(n-1)/2). 至此,可将问题抽象为求最小生成树的边权和. 用了prim+邻接矩阵,prim+邻接表+堆,kruscal各写了一遍,只是内存稍有差别 对于所给的特殊的两个相邻点,只需在开始循环前把这两个点加入子树集合并更新它们所到达的点的min…