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道路建设 (Ver. I)

雪域迷影 2022-01-13 阅读 23
数据结构
#include <iostream>
#include <string>
#include <vector>
using namespace std;
bool go_on(int *visit, int n)
{
    for (int i = 0; i < n; i++)
        if (visit[i] == 0)
            return 1;
    return 0;
}
void prim(int **m, int n)
{
    int *visit = new int[n];
    for (int i = 0; i < n; i++)
        visit[i] = 0;
    int q, t1, t2;
    cin >> q;
    while (q--)
    {
        cin >> t1 >> t2;
        m[t1 - 1][t2 - 1] = 0;
        m[t2 - 1][t1 - 1] = 0;
    }
    struct
    {
        int point;
        int lowcost;
    } closedge[n];

    int temp;
    for (int i = 0; i < n; i++)
    {
        closedge[i].lowcost = 5000;
        closedge[i].point = n + 1;
        for (int j = 0; j < n; j++)
        {
            if (visit[j] == 0 && closedge[i].lowcost > m[i][j])
            {
                closedge[i].lowcost = m[i][j];
                closedge[i].point = j;
            }
        }
    }
    int mmin, loc;
    int last = 0;
    visit[0] = 1;
    while (go_on(visit, n))
    {
        mmin = 5000;
        for (int j = 0; j < n; j++)
        {
            if (visit[j] == 0)
                continue;
            else
            {
                if (mmin > closedge[j].lowcost && closedge[j].point != n + 1)
                {
                    mmin = closedge[j].lowcost;
                    loc = j;
                }
            }
        }
        last += mmin;
        visit[closedge[loc].point] = 1;
        m[loc][closedge[loc].point] = 10000;
        m[closedge[loc].point][loc] = 10000;
        for (int i = 0; i < n; i++)
        {
            closedge[i].lowcost = 5000;
            closedge[i].point = n + 1;
            for (int j = 0; j < n; j++)
            {
                if (visit[j] == 0 && closedge[i].lowcost > m[i][j])
                {
                    closedge[i].lowcost = m[i][j];
                    closedge[i].point = j;
                }
            }
        }
    }
    cout << last << endl;
    delete[] visit;
}
void test(int n)
{
    int k;
    int **m;
    m = new int *[n];
    for (int i = 0; i < n; i++)
    {
        m[i] = new int[n];
        for (int j = 0; j < n; j++)
        {
            cin >> m[i][j];
            if (m[i][j] == 0)
                m[i][j] = 100000;
        }
    }
    prim(m, n);
    delete[] m;
}
int main()
{
    int t;
    while (cin >> t)
        test(t);
    return 0;
}
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