POJ 1258 Agri-Net 最小生成树 Kruskal算法

这个题目有点小坑，它包含多组测试用例，而且需要用~scanf来作为循环条件，用scanf != EOF不行。

#include 
#include 
#include 
using namespace std;
struct Edge
{
    int from, to, cost;
    Edge(int from = 0, int to = 0, int cost = 0) : from(from), to(to), cost(cost) {}
};
vector edges;
int N, par[107], ranks[107], mincost[107], inf = 0x3f3f3f3f;
bool compareEdge(const Edge &a, const Edge &b)
{
    return a.cost < b.cost;
}
void input()
{
    int cost = 0;
    for (int i = 1; i <= N; i++)
    {
        for (int j = 1; j <= N; j++)
        {
            scanf("%d", &cost);
            edges.push_back(Edge(i, j, cost));
        }
    }
}
void init()
{
    for (int i = 0; i <= N; i++)
    {
        par[i] = i;
        ranks[i] = 1;
    }
}
int find(int x)
{
    if (x == par[x])
    {
        return par[x];
    }
    else
    {
        return par[x] = find(par[x]);
    }
}
void unite(int x, int y)
{
    x = find(x);
    y = find(y);
    if (x == y)
    {
        return;
    }
    if (ranks[x] < ranks[y])
    {
        par[x] = y;
    }
    else
    {
        par[y] = x;
        if (ranks[x] == ranks[y])
        {
            ranks[x]++;
        }
    }
}
bool same(int x, int y)
{
    return find(x) == find(y);
}
int kruskal(int s)
{
    sort(edges.begin(), edges.end(), compareEdge);
    for (int i = 1; i <= N; i++)
    {
        mincost[i] = inf;
    }
    int ans = 0;
    for (int i = 0; i < edges.size(); i++)
    {
        if (!same(edges[i].from, edges[i].to))
        {
            ans += edges[i].cost;
            unite(edges[i].from, edges[i].to);
        }
    }
    return ans;
}
int main()
{
    while (~scanf("%d", &N))
    {
        input();
        init();
        printf("%d\n", kruskal(1));
        edges.clear();
    }
    return 0;
}