题目转自hdu 1102,题目传送门 题目大意: 输入一个n*n的邻接矩阵,其中i行j列代表从i到j的路径的长度 然后又m条路已经帮你修好了,求最短要修多长的路才能使所有村庄连接 不难看出,这道题就是标准的最小生成树模板,多水啊 解题思路 虽然很水,但本人还是调了近1h才把代码调好...... 下面介绍一下解决最小生成树的两个方法: Prim 和 Kruskal 一,Prim(不懂的点这里) Prim的思想和dijkstra的想法很想(如果不知道dijkstra算法的请点这里) 那么Prim的复…

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…

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…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1102 题意:这道题实际上和hdu 1242 Rescue 非常相似,改变了输入方式之后, 本题实际上更适合用Prim来做. 用Kruscal的话要做一些变化. /*Constructing Roads Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s)…

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…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1102 Constructing Roads Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 27178    Accepted Submission(s): 10340 Problem Description There are N vil…

题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=1102 Constructing Roads Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 21947    Accepted Submission(s): 8448 Problem Description There are N villa…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1102 Constructing Roads Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 14172    Accepted Submission(s): 5402 Problem Description There are N villa…

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) {…

Constructing Roads Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 19579    Accepted Submission(s): 7474 Problem Description There are N villages, which are numbered from 1 to N, and you should…