hdoj1863 畅通工程（Prime || Kruskal）

题目链接

http://acm.hdu.edu.cn/showproblem.php?pid=1863

思路

最小生成树问题，使用Prime算法或者Kruskal算法解决。这题在hdoj1233的基础上增加了判断是否能形成最小生成树的要求。

代码

Prime算法：

#include <algorithm>
#include <cstring>
#include <cstdio>
using namespace std;

const int INF = 0x7fffffff;
const int N = 100 + 10;
int map[N][N];
int dist[N];
int n, m;

void prime()
{
    int min_edge, min_node;
    for (int i = 1; i <= m; i++)
        dist[i] = INF;
    int ans = 0;
    int now = 1;
    for (int i = 1; i < m; i++)
    {
        dist[now] = -1;
        min_edge = INF;
        for (int j = 1; j <= m; j++)
        {
            if (j != now && dist[j] >= 0)
            {
                if (map[now][j] > 0)
                    dist[j] = min(dist[j], map[now][j]);
                if (dist[j] < min_edge)
                {
                    min_edge = dist[j];    //min_edge存储与当前结点相连的最短的边
                    min_node = j;
                }
            }
        }
        if (min_edge == INF)    //不能形成最小生成树
        {
            puts("?");
            return;
        }
        ans += min_edge;
        now = min_node;
    }
    printf("%d
", ans);
}

int main()
{
    //freopen("hdoj1863.txt", "r", stdin);
    while (scanf("%d%d", &n, &m) == 2 && n)
    {
        memset(map, 0, sizeof(map));
        int a, b, d;
        for (int i = 0; i < n; i++)
        {
            scanf("%d%d%d", &a, &b, &d);
            map[a][b] = map[b][a] = d;
        }
        prime();
    }
    return 0;
}

Kruskal算法：

#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
using namespace std;

struct Edge
{
    int a, b, dist;

    Edge() {}
    Edge(int a, int b, int d):a(a),b(b),dist(d){}
    bool operator < (Edge edge)    
    {
        return dist < edge.dist;
    }
};

const int N = 100 + 10;
vector<Edge> v;
int p[N];
int n, m;

int find_root(int x)
{
    if (p[x] == -1)
        return x;
    else return find_root(p[x]);
}

void kruskal()
{
    memset(p, -1, sizeof(p));
    sort(v.begin(), v.end());    //将边按边长从短到长排序
    int ans = 0;
    for (int i = 0; i < v.size(); i++)
    {
        int ra = find_root(v[i].a);
        int rb = find_root(v[i].b);
        if (ra != rb)
        {
            ans += v[i].dist;
            p[ra] = rb;
        }
    }

    int cnt = 0;
    for (int i = 1; i <= m; i++)
        if (p[i] == -1) cnt++;
    if (cnt > 1)    //连通分量个数多于1个，不能形成最小生成树
    {
        puts("?");
        return;
    }
    printf("%d
", ans);
}

int main()
{
    //freopen("hdoj1863.txt", "r", stdin);
    while (scanf("%d%d", &n, &m) == 2 && n)
    {
        v.clear();
        int a, b, d;
        for (int i = 0; i < n; i++)
        {
            scanf("%d%d%d", &a, &b, &d);
            v.push_back(Edge(a, b, d));
        }
        kruskal();
    }
    return 0;
}