Codeforces 1005 F - Berland and the Shortest Paths

F - Berland and the Shortest Paths

思路：

bfs+dfs

首先，bfs找出1到其他点的最短路径大小dis[i]

然后对于2...n中的每个节点u，找到它所能改变的所有前驱（在保证最短路径不变的情况下)，即找到v，使得dis[v] + 1 == dis[u]，并把u和v所连边保存下来

最后就是dfs递归暴力枚举每个点的前驱，然后输出答案

#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pii pair<int, int>
#define piii pair<int,pii>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

const int N = 2e5 + 5;
int n, m, k;
vector<pii>g[N];
vector<int>pre[N];
int dis[N];
bool vis[N];
char s[N];
vector<string>res;
void bfs(int st) {
    queue<pii>q;
    dis[1] = 0;
    vis[1] = true;
    q.push({1, 0});
    while(!q.empty()) {
        pii p = q.front();
        q.pop();
        for (int i = 0; i < g[p.fi].size(); i++) {
            int v = g[p.fi][i].fi;
            if(!vis[v]) {
                vis[v] = true;
                dis[v] = p.se + 1;
                q.push({v, p.se + 1});
            }
        }
    }
}
void dfs(int u) {
    if((int) res.size() >= k) return ;
    if(u > n) {
        res.pb(s+1);
        return ;
    }
    for (int i = 0; i < pre[u].size(); i++) {
        s[pre[u][i]] = '1';
        dfs(u+1);
        s[pre[u][i]] = '0';
    }
}
int main() {
    fio;
    int u, v;
    cin >> n >> m >> k;
    for (int i = 1; i <= m; i++) {
        cin >> u >> v;
        g[u].pb({v, i});
        g[v].pb({u, i});
    }
    bfs(1);
    for (int i = 2; i <= n; i++) {
        for (int j = 0; j < g[i].size(); j++) {
            pii p = g[i][j];
            if(dis[p.fi]+1 == dis[i]) pre[i].pb(p.se);
        }
    }
    for (int i = 1; i <= m; i++) s[i] = '0';
    dfs(2);
    cout << (int)res.size() << endl;
    for (int i = 0; i < res.size(); i++) cout << res[i] << endl;
    return 0;
}