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 = 1005;
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 = 1;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d %d", &n, &m);
        for(int i = 1; i <= n; i++) father[i] = i;
        for(int i = 1; i <= m; i++)
        scanf("%d %d %d", &p[i].x, &p[i].y, &p[i].w);
        sort(p+1, p+1+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 = 1; 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(1) == Find(n)) break; 
            }
        }
        //注意输出格式
        printf("Scenario #%d:\n%d\n\n", kase++, ans);
    }
    return 0;
}