hdu oj1102 Constructing Roads（最小生成树）

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hdu oj1102 Constructing Roads（最小生成树）

Constructing Roads Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 13995 Accepted Submission(s): 5324 Problem Description There are N villages, which are numbered from 1 to N, and you should…

HDU 1102 Constructing Roads (最小生成树)

最小生成树模板(嗯……在kuangbin模板里面抄的……) 最小生成树(prim) /** Prim求MST * 耗费矩阵cost[][],标号从0开始,0~n-1 * 返回最小生成树的权值,返回-1表示原图不连通 */ const int INF = 0x3f3f3f3f; const int MAXN = 110; bool vis[MAXN]; int lowc[MAXN]; int map[MAXN][MAXN]; int Prim(int cost[][MAXN], int n) {…

hdu 1102 Constructing Roads(最小生成树 Prim)

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1102 Problem 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 conne…

(step6.1.4)hdu 1102(Constructing Roads——最小生成树)

题目大意:输入一个整数n,表示村庄的数目.在接下来的n行中,每行有n列,表示村庄i到村庄 j 的距离.(下面会结合样例说明).接着,输入一个整数q,表示已经有q条路修好. 在接下来的q行中,会给出修好的路的起始村庄和结束村庄.. 输入样例说明如下: 解题思路:最小生成树(kruscal算法) 1)以前的题会直接给村庄编号以及村庄距离.而这道题,这是给出村庄的距离矩阵.村庄的编号信息蕴含在矩阵中.这时候的读取方法为: for(i = 1 ; i <= n ; ++i){ for(j = i +…

HDU 1102 Constructing Roads(最小生成树，基础题)

注意标号要减一才为下标,还有已建设的路长可置为0 题目 #define _CRT_SECURE_NO_WARNINGS #include <stdio.h> #include<string.h> #include <malloc.h> #include<stdlib.h> #include<algorithm> #include<iostream> using namespace std; #define M 110 #define…

hdu Constructing Roads (最小生成树)

题目:http://acm.hdu.edu.cn/showproblem.php?pid=1102 /************************************************************************/ /* hdu Constructing Roads 最小生成树题目大意:在N个村子中已经存在部分存在连通,建最少长度的路使得所有的村子连通. 解题思路:已经连通的村子其中间的路径作为0,即修建的时候修建为0耗费,求这些节点的最小生成树. */ /*…

HDU 1102 Constructing Roads， Prim+优先队列

题目链接:HDU 1102 Constructing Roads Constructing Roads Problem 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 conne…