网址:https://vjudge.net/contest/66965#overview 第一题: poj1251 裸最小生成树 #include<iostream> #include<algorithm> #include<cstring> using namespace std; const int maxn=100500; struct node { int x; int y; int w; }a[maxn]; int n,k; char s,v; int cnt…

A & M - Jungle Roads HDU - 1301 题意:字母之间的路,求最小生成树 题解:处理好建边以后就是一个Prime #include<cstdio> #include<iostream> #include<algorithm> #include<cstring> #include<sstream> #include<cmath> #include<stack> #include<cstd…

学习最小生成树已经有一段时间了 做一些比较简单的题还算得心应手..花了三天的时间做完了kuangbin的专题 写一个题解出来记录一下(虽然几乎都是模板题) 做完的感想:有很多地方都要注意 n == 1 注意double 的精度问题 poj 1251 模板题 大写字母减去'A'+1即是它的编号 #include<stdio.h> #include<string.h> #include<algorithm> #include<map> #include<m…

题意:给你一个矩阵M[i][j]表示i到j的距离 求最小生成树 思路:裸最小生成树 prime就可以了 最小生成树专题 AC代码: #include "iostream" #include "string.h" #include "stack" #include "queue" #include "string" #include "vector" #include "set&…

Borg Maze Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 18644   Accepted: 5990 题目链接:http://poj.org/problem?id=3026 Description: The Borg is an immensely powerful race of enhanced humanoids from the delta quadrant of the galaxy. The Bor…

Constructing Roads Time Limit: 2000MS   Memory Limit: 65536K               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…

Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 15664   Accepted: 6865 Description You are a member of the space station engineering team, and are assigned a task in the construction process of the station. You are expected to write a co…

前两天自学了一点点最短路..看起来很简单的样子... 就去kuangbin的专题找了最简单的一道题练手..然后被自己萌萌的三重for循环超时虐的不要不要的~ 松弛虽然会但是用的十分之不熟练... 代码~ #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int tance[1005][1005]; bool vis[1005]; int dis[1005]; int…

kuangbin字符串专题传送门--http://acm.hust.edu.cn/vjudge/contest/view.action?cid=70325#overview 算法模板: KMP: ; ; int a[MAXN],b[MAXM],Next[MAXM]; int n,m; void getNext(int b[],int Next[]) { ,k=-; Next[]=-; ) { ||b[j]==b[k]) //匹配 { j++,k++; Next[j]=k; } else k=Ne…

https://vjudge.net/problem/Aizu-2224 场景嵌入得很好,如果不是再最小生成树专题里,我可能就想不到解法了. 对所有的边(栅栏)求最大生成树,剩下来的长度即解(也就是需要破环的最小边和). #include<iostream> #include<cstdio> #include<queue> #include<cstring> #include<algorithm> #include<cmath> #i…