还是畅通project

                                             Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

                                                                     Total Submission(s): 32795    Accepted Submission(s): 14769

3

1 2 1

1 3 2

2 3 4

4

1 2 1

1 3 4

1 4 1

2 3 3

2 4 2

3 4 5

0

3

5

 

Huge input, scanf is recommended.

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1233

解题思路：裸的最小生成树，克鲁斯卡尔，边权值排序。

代码例如以下：

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

#define inf 1e9

const int maxn=10005;

int fa[maxn];

struct EG

{

	int x,y,w;

}eg[maxn];

void get_fa()

{

 	for(int i=0;i<=maxn;i++)

 		fa[i]=i;

}

int find(int x)

{

	return x==fa[x]?x:find(fa[x]);

}

void Union(int a,int b)

{

	int a1=find(a);

	int b1=find(b);

	if(a1!=b1)

		fa[a1]=b1;

}

int cmp(EG a,EG b)

{

	return a.w<b.w;

}

int main(void)

{

	int a,b,w,n;

	while(scanf("%d",&n)!=EOF&&n)

	{

		get_fa();

		int m=(n*(n-1))/2,ans=0;

		for(int i=0;i<m;i++)

			scanf("%d%d%d",&eg[i].x,&eg[i].y,&eg[i].w);

		sort(eg,eg+m,cmp);

		for(int i=0;i<m;i++)

		{

			if(find(eg[i].x)!=find(eg[i].y))

			{

				Union(eg[i].x,eg[i].y);

				ans+=eg[i].w;

			}

		}

		printf("%d\n",ans);

	}

}