（图论）51NOD 1212 无向图最小生成树

N个点M条边的无向连通图，每条边有一个权值，求该图的最小生成树。

输入

第1行：2个数N,M中间用空格分隔，N为点的数量，M为边的数量。（2 <= N <= 1000, 1 <= M <= 50000)
第2 - M + 1行：每行3个数S E W，分别表示M条边的2个顶点及权值。(1 <= S, E <= N，1 <= W <= 10000)

输出

输出最小生成树的所有边的权值之和。

输入样例

9 14
1 2 4
2 3 8
3 4 7
4 5 9
5 6 10
6 7 2
7 8 1
8 9 7
2 8 11
3 9 2
7 9 6
3 6 4
4 6 14
1 8 8

输出样例

37
解:Kruskal算法（排序后添加边）的两种实现（时间消耗差不多）：

 #include <stdio.h>
 #include <stdlib.h>
 #include <string.h>
 
 typedef struct Graph
 {
     int s,e,w;
 }graph;
 
 int cmp(const void *a,const void *b)
 {
     return (*(graph *)a).w>(*(graph *)b).w?:-;
 }
 
 graph p[];
 int vis[];
 
 int main()
 {
    int n,m;
    while(scanf_s("%d%d",&n,&m)!=EOF)
    {
         int i,j,ans=;
         memset(vis,,n);
         for(i=;i<m;i++) scanf_s("%d%d%d",&p[i].s,&p[i].e,&p[i].w);
         qsort(p,m,sizeof(graph),cmp);
         vis[p[].s]=;
         vis[p[].e]=;
         ans+=p[].w;
         for(i=;i<n-;i++)
         {
             for(j=;j<m;j++)
             {
                 if((vis[p[j].s]==&&vis[p[j].e]==)||(vis[p[j].s]&&vis[p[j].e]))
                     continue;
                 else
                 {
                     vis[p[j].s]=;
                     vis[p[j].e]=;
                     ans+=p[j].w;
                     break;
                 }
             }
         }
         printf("%d\n",ans);
    }
    return ;
 }

 #include<stdio.h>
 #include<stdlib.h>
 #include<malloc.h>
 
 typedef struct GRAPH
 {
     int s, e, w;
 }graph;
 
 int cmp(const void *a, const void *b)
 {
     return ((graph *)a)->w > ((graph *)b)->w ?  : -;
 }
 
 graph p[];
 int *vis[];
 
 int main()
 {
     int n, m;
     while (scanf_s("%d%d", &n, &m) != EOF)
     {
         int ans = ;
         for (int i = ; i < m; ++i) scanf_s("%d%d%d", &p[i].s, &p[i].e, &p[i].w);
         qsort(p, m, sizeof(graph), cmp);
         for (int i = , j = ; i < m; ++i)
         {
             int *tmp = (int *)malloc(sizeof(int));
             if (vis[p[i].e] == NULL)
             {
                 if (vis[p[i].s] == NULL)
                 {
                     *tmp = j++;
                     vis[p[i].e] = vis[p[i].s] = tmp;
                 }
                 else
                 {
                     vis[p[i].e] = vis[p[i].s];
                     free(tmp);
                 }
             }
             else if (vis[p[i].s] == NULL)
             {
                 vis[p[i].s] = vis[p[i].e];
                 free(tmp);
             }
             else
             {
                 int *a, *b;
                 for (a = vis[p[i].s]; *a > ; a = *a);
                 for (b = vis[p[i].e]; *b > ; b = *b);
                 if (*a == *b)
                 {
                     free(tmp);
                     continue;
                 }
                 else
                 {
                     *tmp = *a;
                     *b = *a = tmp;
                 }
             }
             ans += p[i].w;
         }
         printf("%d\n", ans);
     }
 }

Prim算法（排序后添加点）（写的不好，时间消耗比上面的长）：

 #include <stdio.h>
 #include <malloc.h>
 #include <string.h>
 
 #define MIN(a,b) (pt[a]->pr<pt[b]->pr?a:b)
 
 typedef struct POINT
 {
     struct POINT *p;
     int ed,pr;
 }point;
 
 point *pt[];
 int vis[][];
 int main()
 {
     int n, m;
     while (scanf_s("%d%d", &n, &m) != EOF)
     {
         int ans = ;
         while (m--)
         {
             int a, b, c;
             scanf_s("%d%d%d", &a, &b, &c);
             point *tmp1 = (point *)malloc(sizeof(point)), *tmp2 = (point *)malloc(sizeof(point));
             tmp1->ed = b;
             tmp2->ed = a;
             tmp1->pr = tmp2->pr = c;
             if (pt[a] == NULL|| tmp1->pr < pt[a]->pr)
             {
                 tmp1->p = pt[a];
                 pt[a] = tmp1;
             }
             else for (point *tmp0 = pt[a], *tmp = pt[a]; tmp0 != NULL; tmp = tmp->p)
             {
                 if (tmp == NULL||tmp1->pr < tmp->pr)
                 {
                     tmp1->p = tmp0->p;
                     tmp0->p = tmp1;
                     break;
                 }
                 tmp0 = tmp;
             }
             if (pt[b] == NULL || tmp2->pr < pt[b]->pr)
             {
                 tmp2->p = pt[b];
                 pt[b] = tmp2;
             }
             else for (point *tmp0 = pt[a], *tmp = pt[b]; tmp0 != NULL; tmp = tmp->p)
             {
                 if (tmp == NULL || tmp2->pr < tmp->pr)
                 {
                     tmp2->p = tmp0->p;
                     tmp0->p = tmp2;
                     break;
                 }
                 tmp0 = tmp;
             }
         }
         vis[][] = ;
         vis[][] = ;
         for (int i = ; i < n; i++)
         {
             int tmp = ;
             for (int j = ; j <= i; j++)
             {
                 if (pt[vis[][j]] == NULL) continue;
                 else if(vis[][pt[vis[][j]]->ed]!=)
                 {
                     if ( == tmp) tmp = vis[][j];
                     else tmp = MIN(tmp, vis[][j]);
                 }
                 else
                 {
                     pt[vis[][j]] = pt[vis[][j]]->p;
                     j--;
                 }
             }
             vis[][i+] = pt[tmp]->ed;
             vis[][pt[tmp]->ed] = ;
             ans += pt[tmp]->pr;
             pt[tmp] = pt[tmp]->p;
         }
         printf("%d\n", ans);
     }
 }