HDU - 4460 - Friend Chains

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;
}

标签：HDU,const,int,vis,qe,ans,include,Chains,Friend
来源： https://www.cnblogs.com/shuitiangong/p/12386798.html