【PAT甲级】1072 Gas Station(Dijkstra、枚举)-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_50985046/article/details/124220148

本文介绍了一个C++实现的Dijkstra算法，用于寻找在给定的图中，从一组加油站出发到所有房屋的最短路径和最短距离之和。算法考虑了路径限制条件，即所有房屋的距离之和必须小于一个给定范围。程序首先初始化图的边权重，然后通过Dijkstra算法计算最短路径，并找到满足条件的最佳加油站。

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#include <bits/stdc++.h>

using namespace std;

const int N = 1020;
const int INF = 0x3f3f3f3f;
int n, m, S, T, k, r;
int g[N][N], dist[N];
bool st[N];

int get(string str)
{
    if (str[0] == 'G')
        return n + stoi(str.substr(1));
    return stoi(str);
}

void dijkstra(int start, int& minD, int& minSum)
{
    memset(dist, 0x3f, sizeof dist);
    memset(st, false, sizeof st);
    dist[start] = 0;
    for (int i = 0; i < n + k; i++)
    {
        int t = -1;
        for (int j = 1; j <= n + k; j++)
            if (!st[j] && (t == -1 || dist[t] > dist[j]))
                t = j;
        st[t] = true;
        for (int j = 1; j <= n + k; j++)
        {
            if (dist[j] > dist[t] + g[t][j])
                dist[j] = dist[t] + g[t][j];
        }
    }
    for (int i = 1; i <= n; i++)
    {
        if (dist[i] > r)
        {
            minD = -INF;
            return;
        }
    }
    minD = INF, minSum = 0;
    for (int i = 1; i <= n; i++)
    {
        if (dist[i] < minD) minD = dist[i];
        minSum += dist[i];
    }
}

int main()
{
    memset(g, 0x3f, sizeof g);
    cin >> n >> k >> m >> r; // n屋m路k加油站r范围
    for (int i = 0; i < m; i++)
    {
        string a, b;
        int d;
        cin >> a >> b >> d;
        int x = get(a), y = get(b);
        g[x][y] = g[y][x] = d;
    }
    int minDist = 0, minSum = INF, res = -1;
    for (int i = n + 1; i <= n + k; i++)
    {
        int d1, d2; //该加油站与房屋的最短距离和距离之和
        dijkstra(i, d1, d2);
        if (d1 > minDist) minDist = d1, minSum = d2, res = i;
        if (d1 == minDist && d2 < minSum) minSum = d2, res = i;
    }
    if (res == -1) puts("No Solution");
    else printf("G%d\n%.1f %.1f", res - n, double(minDist), double(minSum) / n + 1e-8);
}