HDU 1233 还是畅通工程（最小生成树）

传送门

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

3
5

prim()

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#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

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#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; }