Codeforces 1209F Koala and Notebook

Posted 2022-12-02 cjlhy

Koala and Notebook

首先肯定要把边拆点， 就变成了最短路问题， 最短路相同的字典序要求最小。

所以我们先bfs找出最短路图建边， 然后dfs把字典序从小到达枚举去更新点， 

这样就是字典序最小的。

其实可以一遍bfs就完成的。

#include<bits/stdc++.h>
#define fi first
#define se second
#define mk make_pair
#define PII pair<int, int>

using namespace std;

const int N = (int)1e6 + 7;
const int mod = (int)1e9 + 7;

int n, m, tot;
int d[N], ans[N];
vector<PII> G[N];
vector<int> G2[N][10];
vector<int> V;

void solve(int u, int v, int x) 
    V.clear();
    for(int i = x; i; i /= 10) 
        V.push_back(i % 10);
    
    reverse(V.begin(), V.end());
    int cur = u;
    for(int i = 0; i < V.size(); i++) 
        if(i == (int)V.size() - 1) 
            G[cur].push_back(mk(V[i], v));
        
        else 
            tot++;
            G[cur].push_back(mk(V[i], tot));
            cur = tot;
        
    
    cur = v;
    for(int i = 0; i < V.size(); i++) 
        if(i == (int)V.size() - 1) 
            G[cur].push_back(mk(V[i], u));
        
        else 
            tot++;
            G[cur].push_back(mk(V[i], tot));
            cur = tot;
        
    


void dfs(vector<int> &P) 
    vector<int> V;
    for(int i = 0; i <= 9; i++) 
        V.clear();
        for(auto &u : P) 
            for(auto &v : G2[u][i]) 
                if(~ans[v]) continue;
                ans[v] = (1LL * ans[u] * 10 + i) % mod;
                V.push_back(v);
            
        
        if(V.size()) dfs(V);
    


int main() 
    scanf("%d%d", &n, &m);
    tot = n;
    for(int i = 1; i <= m; i++) 
        int u, v;
        scanf("%d%d", &u, &v);
        solve(u, v, i);
    
    for(int i = 1; i <= tot; i++) 
        d[i] = -1;
        ans[i] = -1;
    
    d[1] = 0;
    queue<int> que;
    que.push(1);
    while(!que.empty()) 
        int u = que.front();
        que.pop();
        for(auto &e : G[u]) 
            int v = e.se, w = e.fi;
            if(d[v] == -1) 
                d[v] = d[u] + 1;
                G2[u][w].push_back(v);
                que.push(v);
            
            else if(d[u] + 1 == d[v]) 
                G2[u][w].push_back(v);
            
        
    
    ans[1] = 0;
    vector<int> V;
    V.push_back(1);
    dfs(V);
    for(int i = 2; i <= n; i++) 
        printf("%d\n", ans[i]);
    
    return 0;


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