PAT 1030 Travel Plan（Dijsktra算法）

题解

这题是Dijsktra算法的变形，更新节点的时候需要考虑花费（相当于有两个权值），当路径相同时花费优先选择花费少的，加一个判断即可

#include <cstdio>
#include <vector>
#include <algorithm>

const int  maxn=505;
const int  INF=100000000;

using namespace std;

struct Edge
{
    int to;
    int weight;
    int cost;
};

vector<Edge> G[maxn];
int speend[maxn], d[maxn],pre[maxn];
int N, M, S, D;
bool vis[maxn] = {false};

void Dijsktra(int tmp);

int main()
{
    scanf("%d %d %d %d",&N,&M,&S,&D);
    fill(d, d+maxn, INF);
    fill(speend, speend+maxn, INF);
    fill(pre, pre+maxn, -1);
    for(int i=0; i<M; i++)
    {
        int v1;
        Edge e1, e2;
        scanf("%d %d %d %d",&v1,&e1.to,&e1.weight,&e1.cost);
        e2.to = v1;
        e2.cost = e1.cost;
        e2.weight = e1.weight;
        G[v1].push_back(e1);
        G[e1.to].push_back(e2);
    }
    Dijsktra(S);
    int path[maxn], i=1;
    path[0]=D;
    while(pre[D] != -1)
    {
        path[i++] = pre[D];
        D = pre[D];
    }
    for(--i; i>=0; i--)
        printf("%d ", path[i]);
    printf("%d %d", d[path[0]], speend[path[0]]);
    return 0;
}

void Dijsktra(int tmp)
{
    d[tmp]=0;
    speend[tmp]=0;
    for(int j=0; j<N; j++)
    {
        int u=-1, MIN=INF;
        for(int i=0; i<N; i++)
        {
            if (vis[i]==false && d[i]<MIN)
            {
                u = i;
                MIN = d[i];
            }
        }
        if (u == -1)
            return;
        vis[u] = true;
        for (int i=0; i<int(G[u].size()); i++)
        {
            int v, weight, cost;
            v = G[u][i].to;
            weight = G[u][i].weight;
            cost = G[u][i].cost;
            if (vis[v] == false)
            {
               if (d[u]+weight < d[v])
                {
                    d[v] = d[u]+weight;
                    pre[v] = u;
                    speend[v] = speend[u]+cost;
                }
                else if (d[u]+weight == d[v])
                {
                    if(speend[u]+cost < speend[v])
                    {
                        speend[v] = speend[u]+cost;
                        pre[v] = u;
                    }
                }
            }
        }
    }
}