POJ1797 Heavy Transportation

解题思路：典型的Kruskal，不能用floyed（会超时），上代码：

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 #define inf 0x3f3f3f3f
 const int maxn = ;
 int father[maxn];
 
 struct node{
     int x, y, w;
 }p[maxn*maxn]; //第一次开maxn，RE了一次
 
 int cmp(node A, node B)
 {
     return A.w > B.w; //权值从大到小排序
 }
 
 int Find(int x)
 {
     return father[x] == x ? x : father[x] = Find(father[x]);
 }
 
 void Union(int x, int y)
 {
     int rootx = Find(x);
     int rooty = Find(y);
     if(rootx != rooty) father[rootx] = rooty;
     return ;
 }
 
 int main()
 {
     int t, n, m, kase = ;
     scanf("%d", &t);
     while(t--)
     {
         scanf("%d %d", &n, &m);
         for(int i = ; i <= n; i++) father[i] = i;
         for(int i = ; i <= m; i++)
         scanf("%d %d %d", &p[i].x, &p[i].y, &p[i].w);
         sort(p+, p++m, cmp); //注意这里是p+1开始
 
         int ans = inf; //初始化ans为最大值
         for(int i = ; i <= m; i++)
         {
             int rootx = Find(p[i].x);
             int rooty = Find(p[i].y);
             if(rootx != rooty)
             {
                 Union(rootx, rooty);
                 //ans保存路径中最小的权值
                 if(p[i].w < ans) ans = p[i].w;
                 //如果1和n已经连接，则直接跳出
                 if(Find() == Find(n)) break;
             }
         }
         //注意输出格式
         printf("Scenario #%d:\n%d\n\n", kase++, ans);
     }
     return ;
 }