Sunny Cup 2003 - Preliminary Round April 20th, 12:00 - 17:00 Problem E: QS Network In the planet w-503 of galaxy cgb, there is a kind of intelligent creature named QS. QScommunicate with each other via networks. If two QS want to get connected, they…

Facebook Hacker Cup 2014 Qualification Round比赛Square Detector题的解题报告.单击这里打开题目链接(国内访问需要那个,你懂的). 原题如下: Square Detector Problem Description You want to write an image detection system that is able to recognize different geometric shapes. In the first ver…

题目地址:http://blog.csdn.net/shiyuankongbu/article/details/10004443 /* 题意:在i前面找回文子串,在i后面找回文子串相互配对,问有几对 DP:很巧妙的从i出发向两头扩展判断是否相同来找回文串 dpr[i] 代表记录从0到i间的回文子串的个数,dpl[i] 代表记录i之后的回文子串个数 两两相乘就行了 详细解释:http://blog.csdn.net/shiyuankongbu/article/details/10004443 */…

边赋以权值的图称为网或带权图,带权图的生成树也是带权的,生成树T各边的权值总和称为该树的权. 最小生成树(MST):权值最小的生成树. 生成树和最小生成树的应用:要连通n个城市需要n-1条边线路.可以把边上的权值解释为线路的造价.则最小生成树表示使其造价最小的生成树. 构造网的最小生成树必须解决下面两个问题: 1.尽可能选取权值小的边,但不能构成回路: 2.选取n-1条恰当的边以连通n个顶点: MST性质:假设G＝(V,E)是一个连通网,U是顶点V的一个非空子集.若(u,v)是一条具有最小权值的…

最小生成树——Kruskal与Prim算法 序: 首先: 啥是最小生成树??? 咳咳... 如图: 在一个有n个点的无向连通图中,选取n-1条边使得这个图变成一棵树.这就叫“生成树”.(如下图) 每个无向连通图都会拥有至少一个生成树. 而在无向连通图中,我们让每一个边都拥有一个边权(就是每个边代表一个值). 而我们在有边权的无向连通图中构造一个生成树,使得这个生成树所用的边的边权之和最小.这个生成树就叫这个无向连通图的最小生成树! 上图这个最小生成树的边权之和为9,是所有生成树中边权之和最小的.…

2014 Qualification Round Solutions 2013年11月25日下午 1:34 ...最简单的一题又有bug...自以为是真是很厉害! 1. Square Detector (20 Points) When facing a problem in a programming contest there are three main things to consider when planning your solution. In order of importanc…

D. Jerry's Protest 题目连接: http://www.codeforces.com/contest/626/problem/D Description Andrew and Jerry are playing a game with Harry as the scorekeeper. The game consists of three rounds. In each round, Andrew and Jerry draw randomly without replaceme…

暴力 A - Orchestra import java.io.*; import java.util.*; public class Main { public static void main(String[] args) { Scanner cin = new Scanner (System.in); int r = cin.nextInt (); int c = cin.nextInt (); int n = cin.nextInt (); int k = cin.nextInt ();…

在家补补题   模拟 A - Robot Sequence #include <bits/stdc++.h> char str[202]; void move(int &x, int &y, char ch) { if (ch == 'U') x--; if (ch == 'D') x++; if (ch == 'L') y--; if (ch == 'R') y++; } int main(void) { int n; scanf ("%d", &…

C. Cd and pwd commands Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/problemset/problem/158/C Description Vasya is writing an operating system shell, and it should have commands for working with directories. To begin with, he dec…