Nyoj 38 Cabling Problem _ (Solution 1 Kruskal algorithm)

Time limit: +Ms | Memory Limit:65535KB Difficulty:4

Descriptive narrative

Nanyang Institute of Technology to carry out the transformation of power lines. Now the headmaster asked the designer to design a wiring method, the wiring must meet the following conditions:
1, all the buildings are for electricity.
2, the use of wire to spend the least

Input

The first line is an integer n indicating that there is n set of test data.

(n<5)
The first line of the test data for each group is two integer v,e.
V indicates the total number of school buildings (v<=500)
In the subsequent E-line, there are three integers per line a,b,c for the assumption that the construction line between A and B costs C (c<=100). (The two buildings assume that the cost is not specified, which means that the direct connection between the two buildings is too expensive or impossible to connect)
In the next 1 lines. There is a v integer in which the number of I represents the cost of wiring from the I building to the external power supply facility. (0<e<v* (V-1)/2)
(The building number starts from 1), because of security problems, only can choose a building connected to the external power supply equipment.

Data guarantees that there is at least one scenario that satisfies the requirements.

Output

each set of test data outputs a positive integer that represents the minimum cost of laying the line that meets the headmaster's requirements.

Example input

14 61 2 102 3 103 1 101 4 12 4 13 4 11 3 5 6

Example output

4

Code such as the following

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm>using namespace std;const int maxn = 505; struct Arcnode{int v1,v2;//v1, V2 indicates the interconnecting building int cost;//cost represents the cost of connecting V1, v2};int FATHER[MAXN],ADD[MAXN ];int V,e,s;bool cmp (const arcnode &LHS, const Arcnode &RHS) {return lhs.cost < rhs.cost;} void Kruskal (Arcnode *node) {int I,j,k,x,y;i=j=0;s=0;while (J<v-1) {x=father[node[i].v1-1];y=father[node[i].v2-1] if (x!=y) {for (k=0;k<v;k++) if (father[k]==y) father[k]=x;s+=node[i].cost;++j;} ++i;}} int main () {int n;int i;scanf ("%d", &n), while (n--) {scanf ("%d%d", &v,&e); Arcnode *node=new arcnode[e];for (i=0;i<e;i++) scanf ("%d%d%d", &node[i].v1,&node[i].v2,&node[i].cost ); sort (node,node+e,cmp); for (i=0;i<v;i++) {scanf ("%d", &add[i]); father[i]=i;} Sort (add,add+v); Kruskal (node);//Use Kruskal algorithm to find the least cost of connected graph printf ("%d\n", S+add[0]);d elete[] node;node=null;} return 0;}

Nyoj 38 Cabling Problem _ (Solution 1 Kruskal algorithm)