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  • HDU 4460 Friend Chains(bfs)

    题目链接:https://vjudge.net/problem/HDU-4460

    题目大意:给你n个点让你找任意两个点之间最大距离的最小值

      用bfs对每个点做起点的情况做一次搜索,找出其中的最大值

    #include<set>
    #include<map>
    #include<stack>
    #include<queue>
    #include<cmath>
    #include<cstdio>
    #include<cctype>
    #include<string>
    #include<vector>
    #include<climits>
    #include<cstring>
    #include<cstdlib>
    #include<iostream>
    #include<algorithm>
    #define endl '\n'
    #define rtl rt<<1
    #define rtr rt<<1|1
    #define lson rt<<1, l, mid
    #define rson rt<<1|1, mid+1, r
    #define maxx(a, b) (a > b ? a : b)
    #define minn(a, b) (a < b ? a : b)
    #define zero(a) memset(a, 0, sizeof(a))
    #define INF(a) memset(a, 0x3f, sizeof(a))
    #define IOS ios::sync_with_stdio(false)
    #define _test printf("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n")
    using namespace std;
    typedef long long ll;
    typedef pair<int, int> P;
    typedef pair<ll, ll> P2;
    const double pi = acos(-1.0);
    const double eps = 1e-7;
    const ll MOD =  1000000007LL;
    const int INF = 0x3f3f3f3f;
    const int _NAN = -0x3f3f3f3f;
    const double EULC = 0.5772156649015328;
    const int NIL = -1;
    template<typename T> void read(T &x){
        x = 0;char ch = getchar();ll f = 1;
        while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();}
        while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f;
    }
    const int maxn = 1e3+10;
    map<string, int> mark;
    vector<int> edge[maxn];
    int vis[maxn];
    // int bfs(int stt) {
    //     zero(vis);
    //     int ans = 0;
    //     queue<P> qe;
    //     qe.emplace(0, stt);
    //     while(!qe.empty()) {
    //         P t = qe.front();
    //         int u = t.second;
    //         qe.pop();
    //         if (vis[u]) continue; //
    //         vis[u] = true;       //标记的地方不一样速度差了3倍
    //         ans = max(ans, t.first);
    //         ++t.first;
    //         for (auto v : edge[u])
    //             if (!vis[v])
    //                 qe.emplace(t.first, v);
    //     }
    //     return ans;
    // }
    int bfs(int stt) {
        zero(vis);
        int ans = 0;
        queue<P> qe;
        vis[stt] = true;
        qe.emplace(0, stt);
        while(!qe.empty()) {
            P t = qe.front();
            int u = t.second;
            qe.pop();
            ans = max(ans, t.first);
            ++t.first;
            for (auto v : edge[u]) {
                if (vis[v]) continue;
                vis[v] = true;
                qe.emplace(t.first, v);
            }
        }
        return ans;
    }
    int main(void) {
        IOS; int n;
        while(cin >> n && n) {
            string s1, s2;
            for (int i = 0; i<n; ++i) {
                cin >> s2;
                mark[s2] = i+1;
            }
            int m;
            cin >> m;
            while(m--) {
                cin >> s1 >> s2;
                edge[mark[s1]].push_back(mark[s2]);
                edge[mark[s2]].push_back(mark[s1]);
            }
            int ans = 0;
            for (int i = 1; i<=n; ++i) {
                if (edge[i].empty()) {
                    ans = -1;
                    break;
                }
                ans = max(ans, bfs(i));
            }
            cout << ans << endl;
            for (int i = 0; i<=n; ++i)
                edge[i].clear();
            mark.clear();
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/shuitiangong/p/12386798.html
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