Constructing Roads Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 2421 Appoint description:  System Crawler  (2015-05-27) Description There are N villages, which are numbered from 1 to N, and y…

Constructing Roads 题目链接: http://acm.hust.edu.cn/vjudge/contest/124434#problem/D 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 villa…

Constructing Roads 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 betw…

Constructing Roads Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 19884   Accepted: 8315 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…

Constructing Roads Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 20889   Accepted: 8817 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…

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):…

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, or the…

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, or there exists a…

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, or there exists a…

题意:要在n个城市之间建造公路,使城市之间能互相联通,告诉每个城市之间建公路的费用,和已经建好的公路,求最小费用. 解法:最小生成树.先把已经建好的边加进去再跑kruskal或者prim什么的. 代码: #include<stdio.h> #include<iostream> #include<algorithm> #include<string> #include<string.h> #include<math.h> #includ…

给一个n个点的完全图 再给你m条道路已经修好 问你还需要修多长的路才能让所有村子互通 将给的m个点的路重新加权值为零的边到边集里 然后求最小生成树 #include<cstdio> #include<iostream> #include<algorithm> #include<cmath> #include<cstring> #include<string> #define cl(a,b) memset(a,b,sizeof(a))…

题意:有几个村庄,要修最短的路,使得这几个村庄连通.但是现在已经有了几条路,求在已有路径上还要修至少多长的路. 分析:用Prim求最小生成树,将已有路径的长度置为0,由于0是最小的长度,所以一定会被Prim选中加入最小生成树. package Map; import java.util.Scanner; /** * Prime */ public class Poj_2421_Prim { static int MAXVEX = 200; static int n, m; static int[…

题意:给出n个村庄之间的距离,再给出已经连通起来了的村庄.求把所有的村庄都连通要修路的长度的最小值. 思路:Kruskal算法 课本代码: //Kruskal算法 #include<iostream> using namespace std; int fa[120]; int get_father(int x){ return fa[x]=fa[x]==x?x:get_father(fa[x]);//判断两个节点是否属于一颗子树(并查集) } int main(){ int n; int p[…

POJ 1488 题目大意:给定一篇文章,将它的左引号转成 ``(1的左边),右引号转成 ''(两个 ' ) 解题思路:水题,设置一个bool变量标记是左引号还是右引号即可 /* POJ 1488 Tex Quotes --- 水题 */ #include <cstdio> #include <cstring> int main() { #ifdef _LOCAL freopen("D:\\input.txt", "r", stdin); #…

http://poj.org/problem?id=3461 直接KMP就好.水题 #include<cstdio> #include<cstring> const int MAXN=10000+10; const int MAXM=1000000+10; char P[MAXN],T[MAXM]; int f[MAXN],n,m,ans; void getFail() { f[0]=f[1]=0; for(int i=1;i<n;i++){ int j = f[i]; wh…

Biorhythms Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 110991   Accepted: 34541 Description Some people believe that there are three cycles in a person's life that start the day he or she is born. These three cycles are the physical,…

487-3279 Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 236746   Accepted: 41288 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 phras…

Hangover Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 99450   Accepted: 48213 Description How far can you make a stack of cards overhang a table? If you have one card, you can create a maximum overhang of half a card length. (We're as…

一.Description Suppose you are reading byte streams from any device, representing IP addresses. Your task is to convert a 32 characters long sequence of '1s' and '0s' (bits) to a dotted decimal format. A dotted decimal format for an IP address is form…

一.Description The king pays his loyal knight in gold coins. On the first day of his service, the knight receives one gold coin. On each of the next two days (the second and third days of service), the knight receives two gold coins. On each of the ne…

http://poj.org/problem?id=2421 Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 24132   Accepted: 10368 Description There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can con…

题意:给你两个数,求所有的数位的积的和. 析:太水了,没的说,可以先输入边算,也可以最后再算,一样.. 代码如下: #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #inclu…

题目:http://poj.org/problem?id=1035 还是暴搜 #include <iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<stack> #include<queue> #include<iomanip> #include<cmath> #include<map> #include&…

再思考一下好的方法,水过,数据太弱! 本来不想传的! #include <iostream> using namespace std; #define MAX 702 /*284K 422MS*/ typedef struct _point { int x; int y; }point; point p[MAX]; bool judge(point a,point b,point c) { return (a.y-b.y)*(c.x-b.x)-(c.y-b.y)*(a.x-b.x); } in…

题目 http://poj.org/problem?id=2002 题意 已知平面内有1000个点,所有点的坐标量级小于20000,求这些点能组成多少个不同的正方形. 思路 如图,将坐标按照升序排列后,首先枚举p1,p2, 并判断p2是否在p1正下方或者左上角(因为每个正方形只有一条最右边或者是右下的边),按照下图计算p3,p4,判断p3,p4是否存在即可. 感想 排序时要注意和左上角这个信息相符,刚写完时用的是左下角,与升序排序不符合,会遗失部分正方形. 代码 #include <cstdio…

题目链接:POJ 2365 Rope Rope Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7488   Accepted: 2624 Description Plotters have barberically hammered N nails into an innocent plane shape, so that one can see now only heads. Moreover, pursuing th…

一.Description Bill and Ted are taking a road trip. But the odometer in their car is broken, so they don't know how many miles they have driven. Fortunately, Bill has a working stopwatch, so they can record their speed and the total time they have dri…

题意:给定一个n*m的矩阵,让你判断有多少个连通块. 析:用DFS搜一下即可. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring&…

题意:给定一个完全由小写字母组成的字符串s,对每个字母比如x(或a,b,c...z),在字符串中添加或者删除它分别需要花费c1['x']和c2['x']的代价,问将给定字符串变成回文串所需要的最少代价为多少. 解法:设d[i][j]表示将字符串中从第i位至第j位变成回文串所需要的代价.若s[i] == s[j],d[i][j] = d[i+1][j-1]:否则的话,有四种处理方法. 对xa.......by,可以将其变为xa......b,yxa.....by,a.......by,xa....…

关于鸽笼原理的知识看我写的另一篇博客 http://blog.csdn.net/u011026968/article/details/11564841 (需要说明的是,我写的代码在有答案时就输出结果了,但OJ也是从文件读入,所以乍一看我的好像在没输入完就有结果了,但OJ不知道,其实我是直接拿poj3370的代码AC的,32MS,O(∩_∩)O) 直接贴代码 #include<cstdio> #include<cstring> using namespace std; #define…