Farmer John has been elected mayor of his town! One of his campaign promises was to bring internet connectivity to all farms in the area. He needs your help, of course.Farmer John ordered a high speed connection for his farm and is going to share his…

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…

给出一个n*m的图,其中m是人,H是房子,.是空地,满足人的个数等于房子数. 现在让每个人都选择一个房子住,每个人只能住一间,每一间只能住一个人. 每个人可以向4个方向移动,每移动一步需要1$,问所有人移动到房子里的最少花费. 其中,n,m<=100,最多有100个人. 最小给用流裸题. 建立一个超级源点,跟每一个房子连一条边,容量为1,花费为0 每一个人与超级汇点连一条边,容量为1,花费为0 一个房子与每一个人建一条边,房子指向人,容量为1,花费用2者的距离(不是直线距离) 然后跑最小费用流算…

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 题意:无向图有N(N <= 1000)个节点,M(M <= 10000)条边:从节点1走到节点N再从N走回来,图中不能走同一条边,且图中可能出现重边,问最短距离之和为多少? 思路:很经典的构图(看题解的);每条原图中的边赋予cap为1,表示只走一次.超级源点s和汇点t分别和起点终点连边,cap为2,这里cap为2就直接限制了只能有两次最大流:同时最大流中以权值限制得到的就是最小费用:很注意的一点就是此题为无向图带权值,建图时每条有向边建成两条即总边数为4*M.由于spfa找最短路是有…

Description Cows are such finicky eaters. Each cow has a preference for certain foods and drinks, and she will consume no others. Farmer John has cooked fabulous meals for his cows, but he forgot to check his menu against their preferences. Although…

Description FJ's cows really hate getting wet so much that the mere thought of getting caught in the rain makes them shake in their hooves. They have decided to put a rain siren on the farm to let them know when rain is approaching. They intend to cr…

Description You are assigned to design network connections between certain points in a wide area. You are given a set of points in the area, and a set of possible routes for the cables that may connect pairs of points. For each possible route between…

用的是prim算法. 我用vector数组,每次求最小的dis时,不需要遍历所有的点,只需要遍历之前加入到vector数组中的点(即dis[v]!=INF的点).但其实时间也差不多,和遍历所有的点的方法都是16ms... #include <iostream> #include <algorithm> #include <string.h> #include <stdio.h> #include <string> #include <que…

#include<iostream> #include<algorithm> using namespace std; const int N=1e5; struct edge{ int a,b,w; }e[N]; bool cmp(edge a,edge b) { return a.w<b.w; } int p[N]; int find(int x) { if(p[x]!=x) p[x]=find(p[x]); return p[x]; } int main() { int…