PAT A1030 Travel Plan(30)

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题意

还是dijkstra最短路径题，路的边权分两个，dist和cost，第一标尺是dist最小，第二标尺是cost最小，此时唯一。

注意

这道题还要输出路径，所以为每个点记录前驱，最后来一个递归输出。

代码

#include <iostream>#include <algorithm>#include <climits>using namespace std;const int Nmax = 500;int N, M, S, D;int dist[Nmax][Nmax];int cost[Nmax][Nmax];int m_dist[Nmax];int m_cost[Nmax];bool s[Nmax];int v[Nmax];void DIJKSTRA(int s0){    fill(m_dist, m_dist + Nmax, INT_MAX);    fill(s, s + Nmax, false);    m_dist[s0] = 0;    m_cost[s0] = 0;    for (int i = 0; i < N; i++)    {        int min = INT_MAX, k;        for (int j = 0; j < N; j++)        {            if (!s[j] && m_dist[j] < min)            {                min = m_dist[j];                k = j;            }        }        s[k] = true;        for (int j = 0; j < N; j++)        {            if (!s[j] && dist[k][j] != INT_MAX)            {                if (m_dist[k] + dist[k][j] < m_dist[j])                {                    m_dist[j] = m_dist[k] + dist[k][j];                    m_cost[j] = m_cost[k] + cost[k][j];                    v[j] = k;                }                else if (m_dist[k] + dist[k][j] == m_dist[j] && m_cost[k] + cost[k][j] < m_cost[j])                {                    m_cost[j] = m_cost[k] + cost[k][j];                    v[j] = k;                }            }        }    }}void PRINT(int d){    if (d == S)    {        cout << d;        return;    }    PRINT(v[d]);    cout << " " << d;}int main(){    cin >> N >> M >> S >> D;    int c1, c2;    fill(dist[0], dist[0] + Nmax*Nmax, INT_MAX);    for (int i = 0; i < M; i++)    {        cin >> c1 >> c2;        cin >> dist[c1][c2];        cin >> cost[c1][c2];        dist[c2][c1] = dist[c1][c2];        cost[c2][c1] = cost[c1][c2];    }    DIJKSTRA(S);    PRINT(D);    cout << " " << m_dist[D] << " " << m_cost[D];    return 0;}

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