题目链接:http://poj.org/problem?id=1789 题目意思:给出 N 行,每行7个字符你,统计所有的 行 与 行 之间的差值(就是相同位置下字母不相同),一个位置不相同就为1,依次累加.问最终的差值最少是多少. 额.....题意我是没看懂啦= =......看懂之后,就转化为最小生成树来做了.这是一个完全图,即每条边与除它之外的所有边都连通.边与边的权值是通过这个差值来算出来的. #include <iostream> #include <cstdio> us…

题目传送门 题意:给出n个长度为7的字符串,一个字符串到另一个的距离为不同的字符数,问所有连通的最小代价是多少 分析:Kuskal/Prim: 先用并查集做,简单好写,然而效率并不高,稠密图应该用Prim而且要用邻接矩阵,邻接表的效率也不高.裸题但题目有点坑爹:( Kruskal: #include <cstdio> #include <cstring> #include <string> #include <algorithm> #include <…

 POJ 1789 -- Truck History Prim求分母的最小.即求最小生成树 #include<iostream> #include<cstring> #include<algorithm> using namespace std; + ; ; int n;//有几个卡车 ]; int d[maxn];//记录编号的数值 int Edge[maxn][maxn]; int dist[maxn]; void prim() { ; //加入源点 dist[]…

题目连接 http://poj.org/problem?id=1789 Truck History Description Advanced Cargo Movement, Ltd. uses trucks of different types. Some trucks are used for vegetable delivery, other for furniture, or for bricks. The company has its own code describing each…

链接: http://poj.org/problem?id=1789 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22010#problem/G Truck History Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 14950   Accepted: 5714 Description Advanced Cargo Movement, Ltd. uses…

题目链接:POJ 1789 Description Advanced Cargo Movement, Ltd. uses trucks of different types. Some trucks are used for vegetable delivery, other for furniture, or for bricks. The company has its own code describing each type of a truck. The code is simply…

点击打开链接 Truck History Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 15235   Accepted: 5842 Description Advanced Cargo Movement, Ltd. uses trucks of different types. Some trucks are used for vegetable delivery, other for furniture, or fo…

Truck History 题目链接: http://acm.hust.edu.cn/vjudge/contest/124434#problem/E Description Advanced Cargo Movement, Ltd. uses trucks of different types. Some trucks are used for vegetable delivery, other for furniture, or for bricks. The company has its…

Truck History Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 21518   Accepted: 8367 Description Advanced Cargo Movement, Ltd. uses trucks of different types. Some trucks are used for vegetable delivery, other for furniture, or for brick…

Truck History Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 19122   Accepted: 7366 Description Advanced Cargo Movement, Ltd. uses trucks of different types. Some trucks are used for vegetable delivery, other for furniture, or for brick…

题目链接:http://poj.org/problem?id=1789 Advanced Cargo Movement, Ltd. uses trucks of different types. Some trucks are used for vegetable delivery, other for furniture, or for bricks. The company has its own code describing each type of a truck. The code…

模板题 题目:http://poj.org/problem?id=1789 题意:有n个型号,每个型号有7个字母代表其型号,每个型号之间的差异是他们字符串中对应字母不同的个数d[ta,tb]代表a,b之间的差异数问1/Σ(to,td)d(to,td)最大值 prime: #include <iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<stack>…

主题链接:http://poj.org/problem?id=1789 思维:一个一个点,每两行之间不懂得字符个数就看做是权值.然后用kruskal算法计算出最小生成树 我写了两个代码一个是用优先队列写的.可是超时啦,不知道为什么.希望有人能够解答.后面用的数组sort排序然后才AC. code: 数组sort排序AC代码: #include<cstdio> #include<queue> #include<algorithm> #include<iostream…

题目链接:http://poj.org/problem?id=1789 大意: 不同字符串相同位置上不同字符的数目和是它们之间的差距.求衍生出全部字符串的最小差距. #include<stdio.h> #include<math.h> #include<algorithm> using namespace std; ; int cnt; int pre[MAXN]; ]; struct Edge { int from, to; int val; }edge[MAXN *…

Description Advanced Cargo Movement, Ltd. uses trucks of different types. Some trucks are used for vegetable delivery, other for furniture, or for bricks. The company has its own code describing each type of a truck. The code is simply a string of ex…

(￣▽￣)" #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<vector> #include<queue> using namespace std; ; const int INF=10e8; int n,k,minn; ]; int c[MAXN][MAXN],lc[MAXN]; bool vis[MAX…

最小生成树,主要是题目比较难懂. #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; +; *+; int Father[Maxn]; struct Edge { int from,to,w; }edge[maxn]; int n,tot; ]; int Find(int x) { if(x!=Father[x]) Fa…

 http://blog.csdn.net/geniusluzh/article/details/6619575 在说Tarjan算法解决桥和边双连通分量问题之前我们先来回顾一下Tarjan算法是如何求解强连通分量的. Tarjan算法在求解强连通分量的时候,通过引入dfs过程中对一个点访问的顺序dfsNum(也就是在访问该点之前已经访问的点的个数)和一个点可以到达的最小的dfsNum的low数组,当我们遇到一个顶点的dfsNum值等于low值,那么该点就是一个强连通分量的根.因为我们在dfs的…

题目大意:给定一个4位素数,一个目标4位素数.每次变换一位,保证变换后依然是素数,求变换到目标素数的最小步数. 解题报告:直接用最短路. 枚举1000-10000所有素数,如果素数A交换一位可以得到素数B,则在AB间加入一条长度为1的双向边. 则题中所求的便是从起点到终点的最短路.使用Dijkstra或SPFA皆可. 当然,纯粹的BFS也是可以的. 用Dijkstra算法A了题目之后,看了一下Discuss,发现了一个新名词,双向BFS. 即从起点和终点同时进行BFS,相遇则求得最短路. 借鉴了…

http://poj.org/problem?id=1789 Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 27597   Accepted: 10731 Description Advanced Cargo Movement, Ltd. uses trucks of different types. Some trucks are used for vegetable delivery, other for furni…

题目地址: http://poj.org/problem?id=2376 题目内容: Cleaning Shifts Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 12226   Accepted: 3187 Description Farmer John is assigning some of his N (1 <= N <= 25,000) cows to do some cleaning chores arou…

题目地址: http://poj.org/problem?id=1611 题目内容: The Suspects Time Limit: 1000MS   Memory Limit: 20000K Total Submissions: 24253   Accepted: 11868 Description Severe acute respiratory syndrome (SARS), an atypical pneumonia of unknown aetiology, was recogni…

题目地址: http://poj.org/problem?id=2524 题目内容: Ubiquitous Religions Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 26119   Accepted: 12859 Description There are so many different religions in the world today that it is difficult to keep tra…

Description Advanced Cargo Movement, Ltd. uses trucks of different types. Some trucks are used for vegetable delivery, other for furniture, or for bricks. The company has its own code describing each type of a truck. The code is simply a string of ex…

    487-3279 Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 228365   Accepted: 39826 Description Businesses like to have memorable telephone numbers. One way to make a telephone number memorable is to have it spell a memorable word or p…

    Exponentiation Time Limit: 500MS   Memory Limit: 10000K Total Submissions: 126980   Accepted: 30980 Description Problems involving the computation of exact values of very large magnitude and precision are common. For example, the computation of t…

题目链接:http://poj.org/problem?id=2389 题目意思:就是大整数乘法. 题目中说每个整数不超过 40 位,是错的!!!要开大点,这里我开到100. 其实大整数乘法还是第一次写 = =.......大整数加法写得比较多.百练也有一条是大整数乘法,链接如下:http://bailian.openjudge.cn/practice/2980/ 一步一步模拟即可,代码就是按这个来写的. 以 835 * 49 为例(亲爱的读者,允许我截图吧) 简直就是神奇呀----- #inc…

题目链接:http://poj.org/problem?id=2421 实际上又是考最小生成树的内容,也是用到kruskal算法.但稍稍有点不同的是,给出一些已连接的边,要在这些边存在的情况下,拓展出最小生成树来. 一般来说,过到这四组数据大体上就能AC了.  1.题目给出的案例数据    2.连接的道路可能把所有的村庄都已经连通了       3.两个村庄给出多次,即连接这两个村庄的道路是重复的!. 2.3这两种情况的数据如下(为了好看,自己出了一组,当然Sample Input 那组也行):…

题目链接:http://poj.org/problem?id=1611 题意:给定n个人和m个群,接下来是m行,每行给出该群内的人数以及这些人所对应的编号.需要统计出跟编号0的人有直接或间接关系的人数总共有多少. 这又是一条并查集的应用.难点是如何统计出与0有关系的总人数,即包含0的集合内元素的总个数.我的方法是用了两次merge,第一次merge单纯地将同一群内的元素连边,当然该群内的元素的祖先有可能是别的群内的元素,连边的规则是大的元素指向小的元素:第二次merge则把第一次筛选出来的集合中…

题目链接:http://poj.org/problem?id=1007 本题属于字符串排序问题.思路很简单,把每行的字符串和该行字符串统计出的字母逆序的总和看成一个结构体.最后把全部行按照这个总和从小到大排序即可. #include <iostream> #include <algorithm> using namespace std; struct DNA { ]; int count; } d[]; int cmp(DNA a, DNA b) { return a.count…