2021 RoboCom 世界机器人开发者大赛-本科组（决赛）

2021 RoboCom 世界机器人开发者大赛-本科组（决赛）

2021 RoboCom 世界机器人开发者大赛-本科组（决赛）

文章目录

​​2021 RoboCom 世界机器人开发者大赛-本科组（决赛）​​

​​1.绿地围栏​​

​​思路​​​​代码​​

​​2.队列插入​​

​​思路×​​​​代码×​​

​​3.账户安全预警​​

​​思路​​​​代码​​

​​4.猛犸不上 Ban​​

​​思路​​​​代码​​

1.绿地围栏

思路

模拟题目，主要是记住最后要把原点加入到目标点当中，不然最后一个测试点过不了。

代码

#includeusing namespace std;#define int long long#define pii pairint n, l, now_l, x, y;vector v;void init(){ cin >> n >> l; for (int i = 1; i <= n; ++i) { if (i & 1) cin>>y; else cin>>x; v.emplace_back(x,y); } v.emplace_back(0,0); //这里很重要，因为要走回原点，这里如果不加上的话，最后一个测试点就会错。}void solve(){ x = y = 0;now_l = l; cout<

2.队列插入

思路×

不太会，每理解大佬的思路，以后有机会补

代码×

3.账户安全预警

输入样例1

24 3 4daohaole@qq.com 218.109.231.1891jiadelaolao@163.com 112.192.203.187chenyuelaolao@zju.edu.cn 112.18.235.143jiadelaolao@163.com 112.192.203.187chenyuelaolao@zju.edu.cn 113.18.235.143jiadelaolao@163.com 111.192.203.187daohaole@qq.com 218.109.231.189chenyuelaolao@zju.edu.cn 111.18.235.1431jiadelaolao@163.com 115.192.203.187daohaole@qq.com 113.189.58.1411jiadelaolao@163.com 111.192.203.187daohaole@qq.com 112.18.58.1451jiadelaolao@163.com 114.192.203.187chenyuelaolao@zju.edu.cn 112.18.235.143daohaole@qq.com 123.89.158.214chenyuelaolao@zju.edu.cn 112.18.235.143youdaohaole@qq.com 218.109.231.189jiadelaolao@163.com 113.192.203.187youdaohaole@qq.com 218.109.231.189jiadelaolao@163.com 114.192.203.187youdaohaole@qq.com 113.189.58.141youdaohaole@qq.com 123.89.158.2141jiadelaolao@163.com 113.192.203.187youdaohaole@qq.com 112.18.58.145

输出样例1

1jiadelaolao@163.com111.192.203.187 1112.192.203.187 1113.192.203.187 1114.192.203.187 1115.192.203.187 1daohaole@qq.com218.109.231.189 2112.18.58.145 1113.189.58.141 1123.89.158.214 1youdaohaole@qq.com218.109.231.189 2112.18.58.145 1113.189.58.141 1123.89.158.214 1

输入样例2

24 5 8daohaole@qq.com 218.109.231.1891jiadelaolao@163.com 112.192.203.187chenyuelaolao@zju.edu.cn 112.18.235.143jiadelaolao@163.com 112.192.203.187chenyuelaolao@zju.edu.cn 113.18.235.143jiadelaolao@163.com 111.192.203.187daohaole@qq.com 218.109.231.189chenyuelaolao@zju.edu.cn 111.18.235.1431jiadelaolao@163.com 115.192.203.187daohaole@qq.com 113.189.58.1411jiadelaolao@163.com 111.192.203.187daohaole@qq.com 112.18.58.1451jiadelaolao@163.com 114.192.203.187chenyuelaolao@zju.edu.cn 112.18.235.143daohaole@qq.com 123.89.158.214chenyuelaolao@zju.edu.cn 112.18.235.143youdaohaole@qq.com 218.109.231.189jiadelaolao@163.com 113.192.203.187youdaohaole@qq.com 218.109.231.189jiadelaolao@163.com 114.192.203.187youdaohaole@qq.com 113.189.58.141youdaohaole@qq.com 123.89.158.2141jiadelaolao@163.com 113.192.203.187youdaohaole@qq.com 112.18.58.145

输出样例2

1jiadelaolao@163.com111.192.203.187 1112.192.203.187 1113.192.203.187 1114.192.203.187 1115.192.203.187 1

思路

嵌套map，用外层map的键表示邮箱，外层map的值也是一个map（内层map），内层map的键表示ip，值表示ip出现的次数，最后遍历一下，按照规定要求排序后输出即可。

代码

#includeusing namespace std;#define int long long#define psi pairint n, Tip, Tlogin;map > mp;struct node{ string email; //邮箱 vector ip; //存ip登录信息 int tip; //ip总数 int tlogin; //登录次数};vector v, v1;bool cmp(const node& a,const node& b){ if (a.tip != b.tip) return a.tip > b.tip; return a.email < b.email;}bool CMP(psi a, psi b){ if (a.second != b.second) return a.second > b.second; return a.first < b.first;}signed main(){ ios::sync_with_stdio(false),cin.tie(0),cout.tie(0); cin>>n>>Tip>>Tlogin; string s, ip; for (int i = 0; i < n; ++i) { cin>>s>>ip; mp[s][ip] += 1; } for (const auto& it : mp) { node a; a.tlogin = 0; a.email = it.first; a.tip = it.second.size(); for (const auto& x : it.second) { a.ip.emplace_back(x.first, x.second); a.tlogin += x.second; } v.push_back(a); } for (auto & i : v) if (i.tip > Tip && i.tlogin > Tlogin) v1.push_back(i); if (!v1.empty()){ sort(v1.begin(),v1.end(),cmp); for (node &it : v1) { sort(it.ip.begin(),it.ip.end(),CMP); cout<

4.猛犸不上 Ban

输入样例

5 6 1 51 2 12 3 23 4 34 1 53 5 44 5 1

输出样例

11 6Lose!

思路

去年比赛的时候硬写了个搜索处理s->s的情况。

这个题目看到一些大佬的题解，还是很容易理解的。

s->t就不用说了，裸的迪杰斯特拉。

s->s的话，这个题目可以拆成s->i + i -> s这两段，但是边不能重复，因此要先将s->i最短路径上面的所有边都标记一下，然后再跑一次s->i（i->s），这次跑的时候就不能用已经被标记过的边了，也是迪杰斯特拉，只不过需要特殊处理判断一下边的使用情况。

代码

#includeusing namespace std;#define int long longint inf;int n,m,s,t; //如题int mp[505][505]; //存边int dis[505]; //表示s到每个点的距离int dis1[505]; //表示删除第一条路径上的边之后的最短距离int vis[505]; //标记每个点是否被访问过int vis1[505][505]; //标记每条边是否被访问过int pre[505]; //表示每个点的前序最优结点void init(){ cin>>n>>m>>s>>t; memset(mp,0x3f,sizeof mp); memset(dis,0x3f,sizeof dis);dis[s] = 0; memset(pre,255,sizeof pre); inf = dis[0]; while (m--){ int from,to,value;cin>>from>>to>>value; mp[from][to] = mp[to][from] = value; }}void dij(){ while (true){ int mn = inf,k = -1; for (int i = 1; i <= n; ++i) if (!vis[i] && dis[i] < mn) mn = dis[i],k = i; if (k == -1) break; vis[k] = true; for (int i = 1; i <= n; ++i) { if (dis[i] > dis[k] + mp[k][i]) dis[i] = dis[k] + mp[k][i],pre[i] = k; } }}void dij1(){ memset(dis1,0x3f,sizeof dis1);dis1[s] = 0; memset(vis,0,sizeof vis); while (true){ int mn = dis1[0],k = -1; for (int i = 1; i <= n; ++i) if (!vis[i] && dis1[i] < mn) mn = dis1[i],k = i; if (k == -1) break; vis[k] = true; for (int i = 1; i <= n; ++i) { if (vis1[k][i]) continue; if (dis1[i] > dis1[k] + mp[k][i]) dis1[i] = dis1[k] + mp[k][i]; } }}int solve(){ //将s->s 拆分成s->i + i->s int ans = inf; for (int i = 1; i <= n; ++i) { //遍历每个中间点i if (i == s) continue; for (int from = pre[i], to = i; from != -1 ; to = from, from = pre[from]) { vis1[from][to] = vis1[to][from] = true; } //将所有的该路径上的边标记为使用过 dij1(); ans = min(ans,dis[i] + dis1[i]); for (int from = pre[i], to = i; from != -1 ; to = from, from = pre[from]) { vis1[from][to] = vis1[to][from] = false; } //去除标记 } return ans;}signed main(){ ios::sync_with_stdio(false),cin.tie(0),cout.tie(0); init(); //输入等初始化操作 dij(); //跑s->t并且存储pre int ans2 = dis[t]; int ans1 = solve(); //处理s->s cout<<(ans1 == inf ? -1 : ans1)<<" "; cout<<(ans2 == inf ? -1 : ans2)<<"\n"; cout<<(ans1 < ans2 ? "Win!" : "Lose!"); return 0;}

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