标签:class blog code com 数据 2014
题目:有一些电脑,电脑间由一些线路连接。现在给你电脑间的连接状态,以及原来选取的网络线路。
在此基础上加入了K条新线路,要从里面选取新的网络,使得电脑都连通并且线路长度和最小。
分析:图论、最小生成树。
说明:每组数据间有空行。
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cmath>
using namespace std;
typedef struct d_node
{
int point1;
int point2;
int weight;
}enode;
enode edge[1000010];
//union_set
int sets[1000010];
int rank[1000010];
void set_inital( int a, int b )
{
for ( int i = a ; i <= b ; ++ i ) {
rank[i] = 0;
sets[i] = i;
}
}
int set_find( int a )
{
if ( a != sets[a] )
sets[a] = set_find( sets[a] );
return sets[a];
}
void set_union( int a, int b )
{
if ( rank[a] < rank[b] )
sets[a] = b;
else {
if ( rank[a] == rank[b] )
rank[a] ++;
sets[b] = a;
}
}
//end_union_set
int cmp_e( enode a, enode b )
{
return a.weight < b.weight;
}
int kruskal( int n, int m )
{
sort( edge, edge+m, cmp_e );
set_inital( 0, n );
int sum = 0;
for ( int i = 0 ; i < m ; ++ i ) {
int A = set_find( edge[i].point1 );
int B = set_find( edge[i].point2 );
if ( A != B ) {
set_union( A, B );
sum += edge[i].weight;
}
}
return sum;
}
int main()
{
int N,K,M,u,v,w,c = 0;
while ( ~scanf("%d",&N) ) {
int old_cost = 0;
for ( int i = 1 ; i < N ; ++ i ) {
scanf("%d%d%d",&u,&v,&w);
old_cost += w;
}
int e_count = 0;
scanf("%d",&K);
for ( int i = 0 ; i < K ; ++ i ) {
scanf("%d%d%d",&u,&v,&w);
edge[e_count].point1 = u;
edge[e_count].point2 = v;
edge[e_count].weight = w;
e_count ++;
}
scanf("%d",&M);
for ( int i = 0 ; i < M ; ++ i ) {
scanf("%d%d%d",&u,&v,&w);
edge[e_count].point1 = u;
edge[e_count].point2 = v;
edge[e_count].weight = w;
e_count ++;
}
if ( c ++ ) printf("\n");
printf("%d\n%d\n",old_cost,kruskal( N, e_count ));
}
return 0;
}
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UVa 908 - Re-connecting Computer Sites
标签:class blog code com 数据 2014
原文地址:http://blog.csdn.net/mobius_strip/article/details/30129043