這個題就是很簡單的最短路問題。這裏記錄下康復訓練的代碼。

spfa最壞時間復雜度是O(VE)，Dijkstra時間復雜度是O(E+VlogV)。

spfa:

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <queue>
#include <stack>
 
using namespace std;
 
const int maxn=505;
const int INF=0x3f3f3f3f;
int N,M,S,D,u,v,d,c;
struct P
{
    int to,dis,cost;
};
vector<P> edge[maxn];
int distan[maxn],co[maxn],path[maxn],vis[maxn];
 
void dijkstra()
{
    memset(distan,INF,sizeof(distan));
    memset(co,INF,sizeof(co));
    queue<int> qu;
    qu.push(S);
    distan[S]=0;
    co[S]=0;
    while(!qu.empty())
    {
        int U=qu.front();
        qu.pop();
        vis[U]=0;
        for(int i=0;i<edge[U].size();i++)
        {
            int V=edge[U][i].to;
            if(distan[V]>distan[U]+edge[U][i].dis)
            {
                distan[V]=distan[U]+edge[U][i].dis;
                co[V]=co[U]+edge[U][i].cost;
                path[V]=U;
                if(!vis[V])
                {
                    vis[V]=1;
                    qu.push(V);
                }
                continue ;
            }
            else if(distan[V]==distan[U]+edge[U][i].dis)
            {
                if(co[V]>co[U]+edge[U][i].cost)
                {
                    co[V]=co[U]+edge[U][i].cost;
                    path[V]=U;
                    if(!vis[V])
                    {
                        vis[V]=1;
                        qu.push(V);
                    }
                }
            }
        }
    }
}
 
void DFS(int x)
{
    if(path[x]==S)
    {
        cout << S << " " << x;
        return ;
    }
    else DFS(path[x]);
    cout << " " << x;
}
 
int main()
{
    scanf("%d%d%d%d",&N,&M,&S,&D);
    for(int i=1;i<=M;i++)
    {
        scanf("%d%d%d%d",&u,&v,&d,&c);
        edge[u].push_back(P{v,d,c});
        edge[v].push_back(P{u,d,c});
    }
    dijkstra();
    DFS(D);
    cout << " " << distan[D] << " " << co[D] << endl;
    return 0;
}

Dijkstra:

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <queue>
#include <stack>
 
using namespace std;
 
const int maxn=505;
const int INF=0x3f3f3f3f;
int N,M,S,D,u,v,d,c;
struct P
{
    int to,dis,cost;
    bool operator<(const P &a)const
    {
        return dis>a.dis;
    }
};
vector<P> edge[maxn];
int distan[maxn],co[maxn],path[maxn],vis[maxn];
 
void dijkstra()
{
    memset(distan,INF,sizeof(distan));
    memset(co,INF,sizeof(co));
    memset(vis,0,sizeof(vis));
    priority_queue<P> qu;
    qu.push(P{S,0});
    distan[S]=0;
    co[S]=0;
    while(!qu.empty())
    {
        P U=qu.top();
        qu.pop();
        if(vis[U.to]) continue;
        vis[U.to]=1;
        for(int i=0;i<edge[U.to].size();i++)
        {
            P V=edge[U.to][i];
            if(distan[V.to]>distan[U.to]+edge[U.to][i].dis)
            {
                distan[V.to]=distan[U.to]+edge[U.to][i].dis;
                co[V.to]=co[U.to]+edge[U.to][i].cost;
                path[V.to]=U.to;
                qu.push(P{V.to,distan[V.to]});
                continue ;
            }
            else if(distan[V.to]==distan[U.to]+edge[U.to][i].dis)
            {
                if(co[V.to]>co[U.to]+edge[U.to][i].cost)
                {
                    co[V.to]=co[U.to]+edge[U.to][i].cost;
                    path[V.to]=U.to;
                    qu.push(P{V.to,distan[V.to]});
                }
            }
        }
    }
}
 
void DFS(int x)
{
    if(path[x]==S)
    {
        cout << S << " " << x;
        return ;
    }
    else DFS(path[x]);
    cout << " " << x;
}
 
int main()
{
    scanf("%d%d%d%d",&N,&M,&S,&D);
    for(int i=1;i<=M;i++)
    {
        scanf("%d%d%d%d",&u,&v,&d,&c);
        edge[u].push_back(P{v,d,c});
        edge[v].push_back(P{u,d,c});
    }
    dijkstra();
    DFS(D);
    cout << " " << distan[D] << " " << co[D] << endl;
    return 0;
}

然而第一次的AC代碼是不加vis數組優化的spfa，這樣會使得時間增加（好久沒寫了就亂寫一通QAQ）

PAT-Travel Plan (30)-Dijkstra和SPFA