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  • Codeforces 1005 F

    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;
    }
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  • 原文地址:https://www.cnblogs.com/widsom/p/9290144.html
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