Writing a daily newspaper is really a good way to urge learning

It's been rough since summer vacation , Change yourself from today

Monday

Concatenation（ String comparison ）

The order of violence can pass ……

Obviously, the difficulty lies in the case of the same prefix , At this time ab and ba that will do

#include<bits/stdc++.h>
#define _for(i, a, b) for(int i = a; i <= b; i++)
#define rep(i, a, b) for(int i = a; i < b; i++)
using namespace std;

const int N = 2e6 + 10;
string a[N];

bool cmp(string x, string y)
{
    int len = min(x.size(), y.size());
    rep(i, 0, len)
        if(x[i] != y[i])
            return x[i] < y[i];
    return x + y < y + x;
}

int main()
{
    int n; scanf("%d", &n);
    _for(i, 1, n) cin >> a[i];

    sort(a + 1, a + n + 1, cmp);
    _for(i, 1, n) cout << a[i]; 
    puts("");

    return 0;
}

Ancestor（ Prefix suffix + Code tricks ）

During the competition, we went to the classified discussion , It's actually complicated

Remove a point from an interval , There are only prefixes and suffixes left , So we can preprocess the prefix lca And suffixes lca

Because you need to write two copies of the same code , So writing a structure can greatly simplify the code

#include<bits/stdc++.h>
#define _for(i, a, b) for(int i = a; i <= b; i++)
#define rep(i, a, b) for(int i = a; i < b; i++)
using namespace std;

const int N = 1e5 + 10;
int k[N], n, m;

struct Tree
{
    int d[N], up[N][20], q[N], h[N], val[N];
    vector<int> g[N];

    void read()
    {
        _for(i, 1, n) scanf("%d", &val[i]);
        _for(i, 2, n)
        {
            int x; scanf("%d", &x);
            g[x].push_back(i);
        }
    }

    void dfs(int u, int fa)
    {
        d[u] = d[fa] + 1;
        up[u][0] = fa;
        _for(j, 1, 19) up[u][j] = up[up[u][j - 1]][j - 1];

        for(int v: g[u])
        {
            if(v == fa) continue;
            dfs(v, u);
        }
    }

    int lca(int u, int v)
    {
        if(!u || !v) return u + v;
        if(d[u] < d[v]) swap(u, v);
        for(int j = 19; j >= 0; j--)
            if(d[up[u][j]] >= d[v])
                u = up[u][j];
        if(u == v) return u;
        for(int j = 19; j >= 0; j--)
            if(up[u][j] != up[v][j])
                u = up[u][j], v = up[v][j];
        return up[u][0];
    }

    void init()
    {
        dfs(1, 0);
        q[1] = k[1];
        _for(i, 2, m) q[i] = lca(q[i - 1], k[i]);
        h[m] = k[m];
        for(int i = m - 1; i >= 1; i--) h[i] = lca(h[i + 1], k[i]);
    }

    int cal(int x)
    {
        return val[lca(q[x - 1], h[x + 1])];
    }

}A, B;

int main()
{
    scanf("%d%d", &n, &m);
    _for(i, 1, m) scanf("%d", &k[i]);
    A.read(); A.init();
    B.read(); B.init();

    int ans = 0;
    _for(i, 1, m)
        if(A.cal(i) > B.cal(i))
            ans++;
    printf("%d\n", ans);
    
    return 0;
}

Journey（0/1bfs）

Look at the edge as a point , The intersection is regarded as a side map , The weight of the edge is only 0 and 1, Use 0/1bfs that will do , and bfs The complexity is the same

Pay attention to some details , Build drawings id When , Use it carefully unordered_map

0/1bfs and bfs The difference is to use double ended queues , The boundary right is 0 Put your point in front , Otherwise, put it in the back

Update with relax operation , Such a point can join the team twice at most . Then you don't need to vis Array

Besides , The point found for the first time is not necessarily optimal , But the first point of leaving the team must be the best

Actually 0/1bfs It's a special dijsktra, It's just that the priority queue has become a double ended queue , Ensure monotony by joining the team .

#include<bits/stdc++.h>
#define _for(i, a, b) for(int i = a; i <= b; i++)
#define rep(i, a, b) for(int i = a; i < b; i++)
using namespace std;

typedef long long ll;
const int N = 5e5 * 8 + 10;
unordered_map<ll, int> mp;
vector<pair<int, int>> g[N];
int d[N], vis[N], n, cnt;

int id(int x, int y)
{
    ll t = 1LL * x * 1e6 + y;
    if(!mp[t]) mp[t] = ++cnt;
    return mp[t];
}

int solve(int st, int dest)
{
    _for(i, 1, cnt) d[i] = 1e9;
    d[st] = 0;
    deque<int> q;                 //0/1bfs Double ended queue 
    q.push_back(st);

    while(!q.empty())
    {
        int u = q.front(); q.pop_front();
        if(u == dest) return d[u];            //0/1bfs in , It is the minimum value when leaving the team , The first encounter is not necessarily the minimum 
        for(auto x: g[u])
        {
            int v = x.first, w = x.second;
            if(d[v] > d[u] + w)               // Slack operation 
            {
                d[v] = d[u] + w;
                if(!w) q.push_front(v);       //0 Lead the team  1 Put the tail of the team 
                else q.push_back(v);
            }
        }
    }
    return -1;
}

int main()
{
    scanf("%d", &n);
    _for(i, 1, n)
    {
        int c[5];
        _for(j, 1, 4) scanf("%d", &c[j]);
        _for(j, 1, 4)
        {
            if(!c[j]) continue;
            g[id(c[j], i)].push_back({id(i, c[j]), 1});
            int ii = j % 4 + 1;
            if(c[ii]) g[id(c[j], i)].push_back({id(i, c[ii]), 0});
            ii = (j + 1) % 4 + 1;
            if(c[ii]) g[id(c[j], i)].push_back({id(i, c[ii]), 1});
            ii = (j + 2) % 4 + 1;
            if(c[ii]) g[id(c[j], i)].push_back({id(i, c[ii]), 1});
        }
    }

    int s1, s2, t1, t2;
    scanf("%d%d%d%d", &s1, &s2, &t1, &t2);
    int st = id(s1, s2), dest = id(t1, t2);
    printf("%d\n", solve(st, dest));

    return 0;
}

This is a direct question dijsktra You can go through

Of course, pay attention to , Joining the team for the first time is not necessarily the best , The first team is the best . Or simply output after the algorithm runs .

#include<bits/stdc++.h>
#define _for(i, a, b) for(int i = a; i <= b; i++)
#define rep(i, a, b) for(int i = a; i < b; i++)
using namespace std;

typedef long long ll;
const int N = 5e5 * 8 + 10;
unordered_map<ll, int> mp;
int d[N], vis[N], n, cnt;
struct node
{
    int v, w;
    bool operator < (const node& rhs) const
    {
        return w > rhs.w;
    }
};
vector<node> g[N];

int id(int x, int y)
{
    ll t = 1LL * x * 1e6 + y;
    if(!mp[t]) mp[t] = ++cnt;
    return mp[t];
}

int solve(int st, int dest)
{
    _for(i, 1, cnt) d[i] = 1e9;
    d[st] = 0;
    priority_queue<node> q;
    q.push({st, d[st]});

    while(!q.empty())
    {
        node x = q.top(); q.pop();
        int u = x.v;
        if(u == dest) return d[u];
        if(d[u] != x.w) continue;

        for(auto x: g[u])
        {
            int v = x.v, w = x.w;
            if(d[v] > d[u] + w)
            {
                d[v] = d[u] + w;
                q.push({v, d[v]});
            }
        }
    }
    return -1;

}

int main()
{
    scanf("%d", &n);
    _for(i, 1, n)
    {
        int c[5];
        _for(j, 1, 4) scanf("%d", &c[j]);
        _for(j, 1, 4)
        {
            if(!c[j]) continue;
            g[id(c[j], i)].push_back({id(i, c[j]), 1});
            int ii = j % 4 + 1;
            if(c[ii]) g[id(c[j], i)].push_back({id(i, c[ii]), 0});
            ii = (j + 1) % 4 + 1;
            if(c[ii]) g[id(c[j], i)].push_back({id(i, c[ii]), 1});
            ii = (j + 2) % 4 + 1;
            if(c[ii]) g[id(c[j], i)].push_back({id(i, c[ii]), 1});
        }
    }

    int s1, s2, t1, t2;
    scanf("%d%d%d%d", &s1, &s2, &t1, &t2);
    int st = id(s1, s2), dest = id(t1, t2);
    printf("%d\n", solve(st, dest));

    return 0;
}