Agri-Net（POJ 1258）（裸最小生成树）（Prime算法+Kruskal算法）

http://acm.hust.edu.cn/vjudge/problem/10756

题意：给你n个点以及n个点中每两个点之间距离，问你连接所有节点的最小路径总长度是多少。

题解：根据最小生成树的定义可知，这道题是一道最小生成树的题。而算最小生成树的题一般有两种算法，一种是Prime，一种是Kruskal，两种都可以。下文分别给出了两种算法的代码实现（套模板）。

思考：

vector<vector> G(110);//存图的邻接表的时候一般都这么存

Prime算法实现代码：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;
const int INF = 1 << 30;

struct Edge
{
    int v;
    int w;
    Edge (int vv = 0, int ww = INF) : v(vv), w(ww) {}
    bool operator <(const Edge& e) const
    {
        return w > e.w;
    }
};

vector <vector  > G(110);//存图的邻接表的时候一般都这么存

int HeapPrim(const vector<vector > & G, int n)
{
    int i, j, k;
    Edge xDist(0, 0);
    priority_queue pq;
    vector<int> vDist(n);
    vector<int> vUsed(n);
    int nDoneNum = 0;
    for (i = 0; i < n; i++) {
        vUsed[i] = 0;
        vDist[i] = INF;
    }
    nDoneNum = 0;
    int nTotalW = 0;
    pq.push(Edge(0, 0));
    while (nDoneNum < n && !pq.empty()) {
        do {
            xDist = pq.top();
            pq.pop();
        } while (vUsed[xDist.v] == 1 && !pq.empty());
        if (vUsed[xDist.v] == 0) {
            nTotalW += xDist.w;
            vUsed[xDist.v] = 1;
            nDoneNum++;
            for (i = 0; i < G[xDist.v].size(); i++) {
                int k = G[xDist.v][i].v;
                if (!vUsed[k]) {
                    int w = G[xDist.v][i].w;
                    if (vDist[k] > w) {
                        vDist[k] = w;
                        pq.push(Edge(k, w));
                    }
                }
            }
        }
    }
    if (nDoneNum < n) return -1;
    return nTotalW;
}

int main()
{
    #ifndef ONLINE_JUDGE
    freopen ("in.txt", "r", stdin);
    #endif // ONLINE_JUDGE
    int n;
    while (scanf ("%d", &n) != EOF) {
        for (int i = 0; i < n; i++) {
            G[i].clear();
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                int w;
                scanf ("%d", &w);
                G[i].push_back(Edge(j, w));
            }
        }
        printf ("%d\n", HeapPrim (G, n));
    }
    return 0;
}

Kruskal算法实现代码：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;

struct Edge
{
    int s, e, w; //起点，终点，权值
    Edge(int ss, int ee, int ww) : s(ss), e(ee), w(ww) {}
    Edge() {}
    bool operator < (const Edge & e1) const {
        return w < e1.w;
    }
};

vector edges;
vector<int> parent;

int GetRoot(int a)
{
    if (parent[a] == a) return a;
    parent[a] = GetRoot (parent[a]);
    return parent[a];
}

void Merge(int a, int b)
{
    int p1 = GetRoot (a);
    int p2 = GetRoot (b);
    if (p1 == p2) return;
    parent[p2] = p1;
}

int main()
{
    int n;
    while (cin >> n) {
        parent.clear();
        edges.clear();
        for (int i = 0; i < n; i++) parent.push_back(i);
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                int w;
                cin >> w;
                edges.push_back(Edge(i, j, w));
            }
        }
        sort (edges.begin(), edges.end());
        int done = 0;
        int totallen = 0;
        for (int i = 0; i < edges.size(); i++) {
            if (GetRoot (edges[i].s) != GetRoot (edges[i].e)) {
                Merge (edges[i].s, edges[i].e);
                done++;
                totallen += edges[i].w;
            }
            if (done == n - 1) break;
        }
        cout << totallen << endl;
    }
    return 0;
}