题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1162 意义:给出一些点,用线问使所有点直接或间接连通,需要多长: 思路:裸最小生成树: 法1: prim算法:MST(Minimum Spanning Tree,最小生成树)问题有两种通用的解法,Prim算法就是其中之一,它是从点的方面考虑构建一颗MST,大致思想是:设图G顶点集合为U,首先任意选择图G中的一点作为起始点a,将该点加入集合V,再从集合U-V中找到另一点b使得点b到V中任意一点的权值最

Eddy's picture Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8797    Accepted Submission(s): 4476 Problem Description Eddy begins to like painting pictures recently ,he is sure of himself to b

HDU 1233(最小生成树 模板) #include <iostream> #include <algorithm> #include <cstdio> using namespace std; typedef struct { int a,b; int v; }node; const int maxn=105; int ans; int father[maxn]; node graph[maxn*(maxn-1)/2]; int Find(int x) { if(f

Eddy's picture 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1162 ——每天在线,欢迎留言谈论. 题目大意: 给你N个点,求把这N个点连在一起的最短总距离. 思路: 假设每两两点之间都有路径,求最小生成树. AC代码:(Java) import java.util.Scanner; import java.math.*; public class Main { public static final int MAXN = 110;

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1162 [题目大意] 给你n个点的坐标,让你找到联通n个点的一种方法.保证联通的线路最短,典型的最小生成树问题. 方法一 . 通过不断找到最小的边来找到终于结果. Kruskal 算法 #include <iostream> #include <algorithm> #include <cstdio> #include <cmath> using namespac

链接: http://acm.hdu.edu.cn/showproblem.php?pid=1162 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8303    Accepted Submission(s): 4214 Problem Description Eddy begins to like painting pictures

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1162 Eddy's picture Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6866    Accepted Submission(s): 3469 Problem Description Eddy begins to like p

题目链接 Problem Description Eddy begins to like painting pictures recently ,he is sure of himself to become a painter.Every day Eddy draws pictures in his small room, and he usually puts out his newest pictures to let his friends appreciate. but the res

Eddy's picture Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6339    Accepted Submission(s): 3179 Problem Description Eddy begins to like painting pictures recently ,he is sure of himself to b

题目链接 Problem Description Eddy begins to like painting pictures recently ,he is sure of himself to become a painter.Every day Eddy draws pictures in his small room, and he usually puts out his newest pictures to let his friends appreciate. but the res

题目 #define _CRT_SECURE_NO_WARNINGS #include <stdio.h> #include<string.h> #include <malloc.h> #include<stdlib.h> #include<algorithm> #include<iostream> #include<math.h> using namespace std; #define M 110 #define in

poj 1251  && hdu 1301 Sample Input 9 //n 结点数A 2 B 12 I 25B 3 C 10 H 40 I 8C 2 D 18 G 55D 1 E 44E 2 F 60 G 38F 0G 1 H 35H 1 I 353A 2 B 10 C 40B 1 C 200Sample Output 21630 prim算法 # include <iostream> # include <cstdio> # include <cstr

1.HDU  1102  Constructing Roads    最小生成树 2.总结: 题意:修路,裸题 (1)kruskal //kruskal #include<iostream> #include<cstring> #include<cmath> #include<queue> #include<algorithm> #include<cstdio> #define max(a,b) a>b?a:b using na

http://acm.hdu.edu.cn/showproblem.php?pid=1026 模拟一个人走迷宫,起点在(0,0)位置,遇到怪兽要和他决斗,决斗时间为那个格子的数字,就是走一个格子花费时间1, 遇到有数字的格子还要花费这个数字大小的时间,输出最后走到(n-1,m-1)的最小时间,还要输出他的路径,'X'是墙,‘.’是可以走的空地 就是这个路径,刚一看题看到那个样例输出吓得我都飞起来了,好多啊! 其实还好啦,广搜,一开始还因为是普通的广搜,后来发现不一样, 为了寻求最小时间,他可以选

题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=1162 Eddy's picture Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 10085    Accepted Submission(s): 5094 Problem Description Eddy begins to like p

Description 老 Jack 有一片农田,以往几年都是靠天吃饭的.但是今年老天格外的不开眼,大旱.所以老 Jack 决定用管道将他的所有相邻的农田全部都串联起来,这样他就可以从远处引水过来进行灌溉了.当老 Jack 买完所有铺设在每块农田内部的管道的时候,老 Jack 遇到了新的难题,因为每一块农田的地势高度都不同,所以要想将两块农田的管道链接,老 Jack 就需要额外再购进跟这两块农田高度差相等长度的管道. 现在给出老 Jack农田的数据,你需要告诉老 Jack 在保证所有农田全部可连

/* 思路:http://blog.csdn.net/string_yi/article/details/12686873 hdu 1814 输出字典序最小的2-sat */ #include<stdio.h> #include<string.h> #include<math.h> #define N 16100 #define NN 210000 struct node { int v,w,next; }bian[NN*2]; int head[N],cnt,yong

还是畅通工程 Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 43080    Accepted Submission(s): 19636 Problem Description 某省调查乡村交通状况,得到的统计表中列出了任意两村庄间的距离.省政府“畅通工程”的目标是使全省任何两个村庄间都可以实现公路交通(但不一定有直接的公路相连,只要能

最小生成树prime版 大致的步骤 首先选取一个到集合最近的点 然后标记起在集合内部 然后更新最短距离 畅通工程再续 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 24846    Accepted Submission(s): 8035 Problem Description 相信大家都听说一个“百岛湖”的地方吧,百岛湖的居民生活在不

hdu 1503 不知道最后怎么输出,因为公共部分只输出一次.有人说回溯输出,感觉好巧妙!其实就是下图,输出的就是那条灰色的路径,但是初始时边界一定要初始化一下,因为最第一列只能向上走,第一行只能向左走.我用1表示向上走,2向左上方走,3向左走. 刚开始输入字符串时有两种方法,直接输入:或从地址后一位输入,即此时数组起始编号为1.直接输入时,dp[i][j]表示的是以s1[i-1],s2[j-1]为结尾LCS,另一种则就是表示以s1[i],s2[j]为结尾的LCS.两者在路径输出时有些差别,以前

Qin Shi Huang's National Road System Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6137    Accepted Submission(s): 2143 Problem Description During the Warring States Period of ancient China(47