Description The island nation of Flatopia is perfectly flat. Unfortunately, Flatopia has no public highways. So the traffic is difficult in Flatopia. The Flatopian government is aware of this problem. They're planning to build some highways so that i…

题目链接 Description The islandnation of Flatopia is perfectly flat. Unfortunately, Flatopia has no publichighways. So the traffic is difficult in Flatopia. The Flatopian government isaware of this problem. They're planning to build some highways so that…

链接: http://poj.org/problem?id=2485 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22010#problem/H Highways Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 18668   Accepted: 8648 Description The island nation of Flatopia is perfect…

链接:poj 2485 题意:输入n个城镇相互之间的距离,输出将n个城镇连通费用最小的方案中修的最长的路的长度 这个也是最小生成树的题,仅仅只是要求的不是最小价值,而是最小生成树中的最大权值.仅仅须要加个推断 比較最小生成树每条边的大小即可 kruskal算法 #include<cstdio> #include<algorithm> using namespace std; int f[510],n,m; struct stu { int a,b,c; }t[20100]; int…

对于终于生成的最小生成树中最长边所连接的两点来说 不存在更短的边使得该两点以不论什么方式联通 对于本题来说 最小生成树中的最长边的边长就是使整个图联通的最长边的边长 由此可知仅仅要对给出城市所抽象出的图做一次最小生成树 去树上的最长边就可以 #include<bits/stdc++.h> using namespace std; int T,n,a,dist[1020],m[1020][1020]; void prim() { bool p[1020]; for(int i=2;i<=n…

题目 //听说听木看懂之后,数据很水,我看看能不能水过 #define _CRT_SECURE_NO_WARNINGS #include<stdio.h> #include<string.h> #include<math.h> #include<algorithm> using namespace std; #define M 510 #define inf 999999999 int mat[M][M]; int prim(int n,int sta) {…

题目连接 http://poj.org/problem?id=2485 Highways Description The island nation of Flatopia is perfectly flat. Unfortunately, Flatopia has no public highways. So the traffic is difficult in Flatopia. The Flatopian government is aware of this problem. They…

Highways Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 23070   Accepted: 6760   Special Judge Description The island nation of Flatopia is perfectly flat. Unfortunately, Flatopia has a very poor system of public highways. The Flatopian…

POJ2395 Out of Hay 寻找最小生成树中最大的边权. 使用 Kruskal 求解,即求选取的第 \(n-1\) 条合法边. 时间复杂度为 \(O(e\log e)\) . #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; const int maxn = 10005; int n, m, tot, f[2005]; struct edge{ int f…

Sample Input 1 //T 3 //n0 990 692 //邻接矩阵990 0 179692 179 0Sample Output 692 prim # include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # include <cmath> # define LL long long using namespace std ;…