2/10 parallel search set +bfs+dfs+ shortest path +spfa queue optimization

Made a good question ：
The longest path + Chain forward star +spfa

#include <bits/stdc++.h>

using namespace std;
const int inf=0x3f3f3f3f;
const int maxn=1e5+5;
int head[maxn],d,p,c,f,s,cnt,pp[maxn],dis[maxn];
bool vis[maxn],flag;
struct node
{
    
    int to,dis,nxt;
}e[maxn];
void add_edge(int from,int to,int w)
{
    
    e[++cnt].to=to;
    e[cnt].dis=w;
    e[cnt].nxt=head[from];
    head[from]=cnt;
}
queue<int>q;
void spfa()
{
    
    dis[s]=d;
    q.push(s);
    vis[s]=1;
    pp[s]++;
    while(!q.empty())
    {
    
        int u=q.front();q.pop();
        vis[u]=0;
        if(++pp[u]>c)
        {
    
            printf("-1\n");
            flag=1;
            return;
        }
        for(int i=head[u];~i;i=e[i].nxt)
        {
    
            int v=e[i].to;
            if(dis[v]<dis[u]+e[i].dis)
            {
    
                dis[v]=dis[u]+e[i].dis;
                if(!vis[v])
                {
    
                    vis[v]=1;
                    q.push(v);
                }
            }
        }
    }
}
int main()
{
    
    scanf("%d%d%d%d%d",&d,&p,&c,&f,&s);
    head[0]=-1;
    for(int i=1;i<=c;i++)
        head[i]=-1;
    for(int i=1;i<=p;i++)
    {
    
        int u,v;
        scanf("%d%d",&u,&v);
        add_edge(u,v,d);
    }
    for(int i=1;i<=f;i++)
    {
    
        int u,v,w;scanf("%d%d%d",&u,&v,&w);
        add_edge(u,v,d-w);
    }
    spfa();
    if(flag)
        return 0;
    int tmp=0;
    for(int i=1;i<=c;i++)
    {
    
        tmp=max(tmp,dis[i]);
    }
    cout<<tmp<<endl;
    return 0;
}

https://www.luogu.com.cn/problem/P4826
A maximum spanning tree problem , I don't understand the explanation . The key is the following processing , The object of joint search is undirected edge , Therefore, there is no need to build edges in both directions like the shortest path .

for(int i=1;i<=n;i++)
    {
    
        for(int j=i+1;j<=n;j++)
        {
    
            if(i!=j)
                add_edge(i,j,a[i]^a[j]);
        }
    }

#include <bits/stdc++.h>

using namespace std;
const int inf=0x3f3f3f3f;
const int maxn=5e6+5;
int n,m,a[2005],f[2005],minn=inf,cnt,g;
struct node
{
    
    int l,r,w;
}e[maxn];
void add_edge(int from,int to,int w)
{
    
    e[++cnt].l=from;
    e[cnt].r=to;
    e[cnt].w=w;
}
bool cmp(node e1,node e2)
{
    
    return e1.w>e2.w;
}
int r_find(int r)
{
    
    while(r==f[r])
        return r;
    return f[r]=r_find(f[r]);
}
void kruskal()
{
    
    long long sum=0;
    for(int i=1;i<=cnt;i++)
    {
    
        int fx=r_find(e[i].l),fy=r_find(e[i].r);
        if(fx==fy)
            continue;
        g++;
        f[fx]=fy;
        sum+=e[i].w;
        if(g==n-1)
            break;
    }
    cout<<sum<<endl;
}
int main()
{
    
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    for(int i=1;i<=n;i++)
        f[i]=i;
    for(int i=1;i<=n;i++)
    {
    
        for(int j=i+1;j<=n;j++)
        {
    
            if(i!=j)
                add_edge(i,j,a[i]^a[j]);
        }
    }
    sort(e+1,e+cnt+1,cmp);
    kruskal();
    return 0;
}

2.bfs
mp Record the points on the chart and arrive in several steps
f[100005][3]： Look for the a,b Data range of , Here is how many points exist
bool vis[maxn][maxn]： Judge whether this point has passed , There's a caveat , Reaching this point first means that it takes the least time

#include <bits/stdc++.h>

using namespace std;
const int inf=0x3f3f3f3f;
const int maxn=5050;
int mp[maxn][maxn],f[100005][3],n,m,a,b;
bool vis[maxn][maxn];
int dx[4]={
    1,-1,0,0};
int dy[4]={
    0,0,-1,1};
struct node
{
    
    int x,y,step;
};
queue<node>q;
void tag(int x,int y)
{
    
    node cur;
    cur.x=x;
    cur.y=y;cur.step=0;
    q.push(cur);
    vis[x][y]=1;
    return;
}
void bfs()
{
    
    node cur,nxt;
    while(!q.empty())  // Not empty return 0
    {
    
        cur=q.front();
        mp[cur.x][cur.y]=cur.step;
        q.pop();
        for(int i=0;i<4;i++)
        {
    
            int x=cur.x+dx[i];
            int y=cur.y+dy[i];
            if(x>=1&&x<=n&&y>=1&&y<=n&&!vis[x][y])
            {
    
                vis[x][y]=1;
                nxt.x=x;nxt.y=y;
                nxt.step=cur.step+1;
                q.push(nxt);
            }
        }

    }
}
int main()
{
    
    scanf("%d%d%d%d",&n,&m,&a,&b);
    for(int i=1;i<=a;i++)
    {
    
        int x,y;scanf("%d%d",&x,&y);
        tag(x,y);
    }
    for(int i=1;i<=b;i++)
    {
    
        scanf("%d%d",&f[i][1],&f[i][2]);
    }
    bfs();
    for(int i=1;i<=b;i++)
    {
    
        printf("%d\n",mp[f[i][1]][f[i][2]]);
    }
    return 0;
}

3.https://www.luogu.com.cn/problem/P2196
A very classic deep search topic .
From the transformation of state ：
If there is a road connected with the current floor and it has not been visited , It means that you can always dig down ;
When the digging goes on , Just record the path , Sum maximum , Go back ;
The importance of backtracking is needless to say

#include <bits/stdc++.h>

using namespace std;
const int inf=0x3f3f3f3f;
const int maxn=50;
int a[maxn],f[maxn][maxn],n,m,b,path[maxn],tmp,ans[maxn],gg;
bool vis[maxn];
int check(int x)
{
    
    for(int i=1;i<=n;i++)
    {
    
        if(f[x][i]&&!vis[i])  // It can also dig down and return 1
            return 1;
    }
    return 0; // I can't dig any more and return 0
}
void dfs(int x,int step,int sum)
{
    
    if(!check(x))
    {
    
        if(tmp<sum)
        {
    
            tmp=sum;gg=step;
            for(int i=1;i<=step;i++)
                ans[i]=path[i];
        }
        return;
    }
    for(int i=1;i<=n;i++)
    {
    
        if(f[x][i]&&!vis[i])
        {
    
            vis[i]=1;
            path[step+1]=i;
            dfs(i,step+1,sum+a[i]);
            vis[i]=0;
        }
    }
}
int main()
{
    
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    for(int i=1;i<=n;i++)
    {
    
        for(int j=i+1;j<=n;j++)
        {
    
            int x;scanf("%d",&x);
            f[i][j]=x;
        }
    }
    for(int i=1;i<=n;i++)
    {
    
        path[1]=i;
        vis[i]=1;
        dfs(i,1,a[i]);
        vis[i]=0;
    }
    for(int i=1;i<=gg;i++)
    {
    
        cout<<ans[i]<<" ";
    }
    cout<<endl;
    cout<<tmp<<endl;
    return 0;
}

Graph storage of chained forward stars + Deep search

#include <bits/stdc++.h>

using namespace std;
const int maxn=2e5+5;
struct node
{
    
    int to,dis,nxt;
}e[maxn];
int head[maxn],f[maxn],n,m,cnt;
bool vis[maxn];
void add_edge(int from,int to,int dis)
{
    
    e[++cnt].to=to;
    e[cnt].dis=dis;
    e[cnt].nxt=head[from];
    head[from]=cnt;
}
void dfs(int x,int val)
{
    
    f[x]=val;
    vis[x]=1;
    for(int i=head[x];~i;i=e[i].nxt)
    {
    
        if(!vis[e[i].to])
            dfs(e[i].to,val^e[i].dis);
    }
}
int main()
{
    
    scanf("%d",&n);
    head[0]=-1;
    for(int i=1;i<=n;i++)
        head[i]=-1;
    for(int i=1;i<=n-1;i++)
    {
    
        int u,v,w;scanf("%d%d%d",&u,&v,&w);
        add_edge(u,v,w);add_edge(v,u,w);
    }
    dfs(1,0);
    scanf("%d",&m);
    for(int i=1;i<=m;i++)
    {
    
        int x,y;
        scanf("%d%d",&x,&y);
        cout<<(f[x]^f[y])<<endl;
    }
    return 0;
}