链接：

2737. 找到最近的标记节点

题意：

给一个n节点有向图，求节点s和点集marked中的最短距离，没有可以的到达的点则返回-1

解：

摇了一题困难过了2/3，搞不定了

没有负权边的单源最短距离，迪杰斯特拉

跑一组WA了（忘记最大改最小了）的案例结果执行出错了，怎么肥四啊力扣

实际代码：

#include<bits/stdc++.h>
using namespace std;
const int Nmax=5E2+7;
int minimumDistance(int n, vector<vector<int>>& edges, int s, vector<int>& marked)
{
    vector<vector<int>>mps(n,vector<int>(n,INT_MAX));
    for(auto &edge:edges)
    {
        mps[edge[0]][edge[1]]=min( mps[edge[0]][edge[1]],edge[2] );
        //新图 去重 
    }
    vector<int>distence(n,INT_MAX);distence[s]=0;
    vector<bool>book(n);
    int now=s,min_index,min_num=INT_MAX,ans=INT_MAX;
    while(true)
    {
        book[now]=1;min_num=INT_MAX;
        //distence[now]=0;//标记选取 
        auto &mp=mps[now];//遍历邻接点 
        for(int index=0;index<n;index++)
        {
            if(index!=now && mp[index]!=INT_MAX)//不是当前点且有边 
            {
                distence[index]=min(distence[index],distence[now]+mp[index]);//更新距离 
            }
            
            if(book[index]==0 && distence[index]<min_num)//选取最小的未选取的点 
            {
                min_num=distence[index];
                min_index=index;
            }
        }
        if(min_num!=INT_MAX) now=min_index;
        else break;
    }
    for(auto m:marked)
    {
        //cout<<m<<" "<<distence[m]<<endl;
        ans=min(ans,distence[m]);
    }
    return (ans==INT_MAX?-1:ans);
}
int main()
{
    vector<vector<int>> edges;
    vector<int> marked;
    cout<<"输入n和edges.size"<<endl;
    int n,s,temp;cin>>n>>s;
    for(int i=1;i<=s;i++)
    {
        cout<<"输入edge"<<endl;
        int a,b,c;cin>>a>>b>>c;
        vector<int>temp{a,b,c};
        edges.push_back(temp);
    }
    cout<<"输入s和marked,输入EOF结束"<<endl;
    cin>>s;
    while(cin>>temp)
    {
        marked.push_back(temp);
    }
    
    int ans=minimumDistance(n,edges,s,marked);
    cout<<ans<<endl;
    return 0;
}

限制：

• 2 <= n <= 500
• 1 <= edges.length <= 104
• edges[i].length = 3
• 0 <= edges[i][0], edges[i][1] <= n - 1
• 1 <= edges[i][2] <= 106
• 1 <= marked.length <= n - 1
• 0 <= s, marked[i] <= n - 1
• s != marked[i]