还是畅通工程

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 41447 Accepted Submission(s): 18920

Description

某省调查乡村交通状况，得到的统计表中列出了任意两村庄间的距离。省政府“畅通工程”的目标是使全省任何两个村庄间都可以实现公路交通（但不一定有直接的公路相连，只要能间接通过公路可达即可），并要求铺设的公路总长度为最小。请计算最小的公路总长度。

Input

测试输入包含若干测试用例。每个测试用例的第1行给出村庄数目N ( < 100 )；随后的N(N-1)/2行对应村庄间的距离，每行给出一对正整数，分别是两个村庄的编号，以及此两村庄间的距离。为简单起见，村庄从1到N编号。
当N为0时，输入结束，该用例不被处理。

Output

对每个测试用例，在1行里输出最小的公路总长度。

Sample Input

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

Sample Output

prim()

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int maxn = 105;
const int INF = 0x3f3f3f3f;
struct Edge{
	int u,v,w,next;
	bool operator < (const Edge &x)const
	{
		return w > x.w;
	}
}edge[maxn*(maxn-1)];
int tot = 0,head[maxn],dis[maxn];
bool vis[maxn];

void init()
{
	tot = 0;
	memset(head,-1,sizeof(head));
}

void addedge(int u,int v,int w)
{
	edge[tot] = (Edge){u,v,w,head[u]
	};
	head[u] = tot++;
}

int prim()
{
	int sum = 0;
	priority_queue<Edge>que;
	Edge p;
	memset(dis,INF,sizeof(dis));
	memset(vis,false,sizeof(vis));
	p.v = 1;
	que.push(p);
	dis[1] = 0;
	while (!que.empty())
	{
		p = que.top();
		que.pop();
		int u = p.v;
		if (vis[u])	continue;
		vis[u] = true;
		sum += dis[u];
		for (int i = head[u];~i;i = edge[i].next)
		{
			int v = edge[i].v,w = edge[i].w;
			if (dis[v] > w)
			{
				dis[v] = edge[i].w;
				p.u = u,p.v = v,p.w = w;
				que.push(p);
			}
		}
	}
	return sum;
}

int main()
{
	//freopen("input.txt","r",stdin);
	int N;
	while (~scanf("%d",&N) && N)
	{
		int u,v,w;
		int len = N*(N - 1)/2;
		init();
		for (int i = 0;i < len;i++)
		{
			scanf("%d%d%d",&u,&v,&w);
			addedge(u,v,w);
			addedge(v,u,w);
		}
		printf("%d\n",prim());
	}
	return 0;
}

Kruskal

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef __int64 LL;
const int maxn =  105;
struct Edge{
	int u,v,next;
	LL w;
}edge[maxn*(maxn-1)<<1];
int tot = 0,head[maxn],fa[maxn];

void init(int N)
{
	tot = 0;
	memset(head,-1,sizeof(head));
	for (int i = 0;i <= N;i++)	fa[i] = i;
}

bool cmp(struct Edge x,struct Edge y)
{
	return x.w < y.w;
}

int find(int x)
{
	int r = x;
	while (r != fa[r])	r = fa[r];
	int i = x,j;
	while (i != r)
	{
		j = fa[i];
		fa[i] = r;
		i = j;
	}
	return r;
}

int main()
{
	//freopen("input.txt","r",stdin);
	int N;
	while (~scanf("%d",&N) && N)
	{
		LL sum = 0;
		int len = N*(N-1)/2;
		init(N);
		for (int i = 0;i < len;i++)
		{
			scanf("%d%d%I64d",&edge[i].u,&edge[i].v,&edge[i].w);
		}
		sort(edge,edge+len,cmp);
		for (int i = 0;i < len;i++)
		{
			int fx = find(edge[i].u),fy = find(edge[i].v);
			if (fx != fy)
			{
				fa[fx] = fy;
				sum += edge[i].w;
			}
		}
		printf("%I64d\n",sum);
	}
	return 0;
}