Timus 1982 Electrification Plan (minimum spanning tree)

Electrification Plantime limit:0.5 Second
Memory limit:64 Mbsome Country has NCities. The government have decided to electrify all these cities. At first, power stations in kDifferent cities were built. The other cities should is connected with the power stations via power lines. For any cities I, JIt's possible to Builda power line between them in C IJRoubles. The country is in crisis after a civil war and so the government decided to build has only a few power lines. Of course from every city there must is a path along the lines to some city with a power station. Find the minimum possible cost-to-build all necessary power lines. Inputthe First line contains integers Nand k(1≤ k≤ N≤100). The second line contains kDifferent integers that is the numbers of the cities with power stations. The next NLines contain an NX NTable of integers { C IJ} (0≤ C IJ≤105). It's Guaranteed that C IJ= C ji, C IJ> 0 for I≠ J, C II= 0.OutputOutput the minimum cost to electrify all the cities. Sample

input	Output
4 21 40 2 4 32 0 5 24 5 0 13 2 1 0	3

problem Author:Mikhail Rubinchik "Analysis" assigns the distance between any two cities with a power station to 0, and then prim the minimum spanning tree. (Such a simple idea I actually did not think, or the seniors taught me).

#include <iostream>#include<cstring>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<time.h>#include<string>#include<map>#include<stack>#include<vector>#include<Set>#include<queue>#defineINF 10000000#defineMoD 10000typedefLong Longll;using namespacestd;Const intn=6005;Const intm=50000;intPowerintAintBintc) {intans=1; while(b) {if(b%2==1) {ans= (ans*a)%c;b--;} B/=2; a=a*a%c;}returnans;}intA[n],w[n][n],vis[n],lowcost[n];intn,m,k;voidPrim () {intsum=0; lowcost[1]=-1;  for(intI=2; i<=n;i++) {Lowcost[i]=w[1][i]; }     for(intI=1; i<n;i++){        intminn=inf,k;  for(intj=1; j<=n;j++){            if(lowcost[j]!=-1&&lowcost[j]<Minn) {k=j;minn=Lowcost[j]; }} sum+=Minn; LOWCOST[K]=-1;  for(intj=1; j<=n;j++) {Lowcost[j]=min (lowcost[j],w[k][j]); }} printf ("%d\n", sum);}intMain () {scanf ("%d%d",&n,&m);  for(intI=1; i<=m;i++) scanf ("%d",&A[i]);  for(intI=1; i<=n;i++) for(intj=1; j<=n;j++) scanf ("%d",&W[i][j]);  for(intI=1; i<=m;i++) for(intj=1; j<=m;j++) w[a[i]][a[j]]=0;    Prim (); return 0;}

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Timus 1982 Electrification Plan (minimum spanning tree)