還是暢通工程
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 49125 Accepted Submission(s): 22415
當N爲0時,輸入結束,該用例不被處理。
//代碼如下:
#include <stdio.h>
#include <algorithm>
using namespace std;
const int maxn = 102;
int par[maxn];
int n;
struct Road
{
int st;
int endd;
int value;
}p[5050];
void init(int n)
{
for (int i = 0; i <= n; i++)
par[i] = i;
}
bool cmp(Road a,Road b)
{
return a.value < b.value;
}
int find(int x)
{
if (x != par[x])
return par[x] = find(par[x]);
return x;
}
void unite(int x,int y)
{
int fa = find(x);
int fb = find(y);
if (fa != fb)
{
par[fb] = fa;
}
}
int main()
{
int n;
while (~scanf("%d",&n) && n)
{
init(n);
int t = n * (n - 1) / 2;
for (int i = 0; i < t; i++)
{
scanf ("%d%d%d",&p[i].st,&p[i].endd,&p[i].value);
}
sort(p,p+t,cmp);
int sum = 0;
for (int i = 0; i < t; i++)
{
if (find(p[i].st) != find(p[i].endd))
{
unite (p[i].st,p[i].endd);
sum +=p[i].value;
}
}
printf ("%d\n",sum);
}
return 0;
}