2025年睿抗机器人开发者大赛CAIP-编程技能赛(省赛)-RoboCom 世界机器人开发者大赛-本科组
RC-u1 早鸟价
代码
#include "bits/stdc++.h" using namespace std; int main(){int n;cin>>n;int m, d, q;while(n--){cin>>m>>d>>q;if(m > 7 || m == 7 && d > 11) {cout<<"Too late!"; }else if(m < 6 || m == 6 && d <= 20) {if(q < 1800) cout<<"Need more!";else if(q == 1800) cout<<"Ok!" ;else cout<<"^_^"; }else {if(q < 2000) cout<<"Need more!";else if(q == 2000) cout<<"Ok!" ;else cout<<"^_^";}cout<<endl;}return 0; }
RC-u2 谁进线下了？III
代码：
#include "bits/stdc++.h" using namespace std; int main(){int T;cin>>T;int n, s;int cnt = 0;int p, f;int summ = 0;while(T--){cin>>n>>s;int m = n;cnt = 0;summ = 0;while(m--){cin>>p>>f;if(p == 1) cnt ++;summ += f;}if(cnt >= n* 1.0 / 2) cout<<"1";else cout<<"0";cout<<" ";if(summ - s >= 50) cout<<"1";else cout<<"0";cout<<endl;}return 0; }
RC-u3 点格棋
思路：这题是个模拟，要注意的是判断条件，一是边界，二是下棋人的顺序，三是是否是相邻的格子，然后是连过的不能连，要注意的时候有可能一次连线拿两分，还有就是map如果结构体当键，无论是否里面的数据一样的，都会算成不同的键
代码：#include "bits/stdc++.h" using namespace std; int n, m, s; int mp[102][102][102][102] = {0}; int check(int x1, int y1, int x2, int y2){int score = 0;if(x1 == x2){if(x1 - 1 >= 1){if(mp[x1 - 1][y1][x1][y1] && mp[x1 - 1][y2][x1][y2] && mp[x1 - 1][y1][x1 - 1][y2])score++; }if(x1 + 1 <= n ){if(mp[x1 + 1][y1][x1][y1] && mp[x1 + 1][y2][x1][y2] && mp[x1 + 1][y1][x1 + 1][y2] )score++; } }else if(y1 == y2){if(y1 - 1 >= 1){if(mp[x1][y1- 1][x1][y1] && mp[x1 ][y1 - 1][x2][y2- 1] && mp[x2][y2 - 1][x2][y2] )score++;}if(y1 + 1 <= m ){if(mp[x1][y1 + 1][x1][y1] && mp[x1 ][y1 + 1][x2][y2+ 1] && mp[x2][y2 + 1][x2][y2])score++;}}return score; } int main(){cin>>n>>m>>s;int a, x1, y1, x2, y2; vector<int> v;int cnt = 0;int A = 0, B = 0;int next = 0;for(int i = 1; i <= s; i ++){cin>>a>>x1>>y1>>x2>>y2;if(a == next && x1 >= 1 && y1 >= 1 && x2 >= 1 && y2 >= 1 && x1 <= n && x2 <= n && y1 <= m && y2 <= m&&(abs(x1 - x2 )+abs(y1 - y2) == 1)&&!mp[x1][y1][x2][y2] && (x1 == x2 || y1 == y2)){mp[x1][y1][x2][y2] = 1;mp[x2][y2][x1][y1] = 1;int f = check(x1, y1, x2, y2);if(f == 0) next = 1 - next;if(a == 0) A += f; else B += f; } else v.push_back(i),cnt ++; }if(cnt == 0) cout<<-1;else for(int i = 0; i < v.size();i ++){cout<<v[i];if(i != v.size() - 1) cout<<" ";} cout<<endl;if(A > B) cout<<0<<" "<<A; else cout<<1<<" "<<B;return 0; }
RC-u4 Tree Tree 的
思路：很明显是找环，一个最大的，一个次大的，但是注意他这个图不一定是连通的，也就是说你从一个点开始dfs可能遍历不到所有的点。我这里用了一个dfs来进行判断，也就是一个dfs找一个连通图中所有的环。
代码：#include "bits/stdc++.h" using namespace std; vector<int> g[15]; int ans1 = 1, ans2 = 0; int pre[60], vis[20], vis1[20]; void dfs(int x, int p){vis[x] = 1;vis1[x] = 1;for(int i = 0; i < g[x].size(); i ++){int t = g[x][i]; vis1[t] = 1;if(vis[t]) {int cnt = 1;int xx = x;while( xx != t){xx = pre[xx];cnt ++;}if(cnt > ans1){ans2 = ans1;ans1 = cnt; }if(cnt < ans1 && cnt > ans2){ans2 = cnt;} } else if(t == p) continue; else if(!vis[t]){pre[t] = x;dfs(t, x);} }vis[x] = 0; } int main(){int T;cin>>T;while(T--){int n, m, x, y;cin>>n>>m;memset(vis, 0, sizeof(vis));memset(vis1, 0, sizeof(vis1));memset(pre, 0, sizeof(pre));for(int i = 0; i < 11; i ++) g[i].clear();while(m--){cin>>x>>y;g[x].push_back(y);g[y].push_back(x);}ans1 = 1;ans2 = 0;for(int i = 1; i <= n; i ++)if(!vis1[i]) dfs(i, -1);cout<<ans1<<" "<<ans2<<endl;}return 0; }
RC-u5 游戏设计师
思路：这题大模拟，耐心做一下就好了，细节比较多，然后的话他这个你注意是1是实心格子，2是易碎格子，直接模拟的话能拿22分，后面两个超时，这时候会想到（我看大家好像都想到了哈哈，本来以为是剪枝但是感觉没有什么好剪得）直接从终点开始遍历就能打表所有的可能了，再多询问也是O(1), 遍历的话直接bfs就可以了，这里注意可能一开始就在终点，这是第1个测试点一分。
代码：
#include "bits/stdc++.h" using namespace std; const int N = 1e3 + 10; int n, m; int g[N][N]; int vis[4][1001][1001]; int ex = -1, ey = -1; struct node{int op;int x, y;int dis;node(int a, int b, int c, int e){op = a, x = b, y = c, dis = e;} }; queue<node> q; void bfs(int op, int x, int y){int x1, y1, x2, y2;if(op == 0)q.push(node(0, x, y, 0));else if(op == 1){q.push(node(1, x, y, 0));}else {q.push(node(2, x, y, 0));}int opp = -1;while(q.size()){node t = q.front();q.pop();int op1 = t.op;if(op1 == 0){x1 = t.x - 1, x2 = t.x - 2, y1 = t.y, y2 = t.y, opp = 2;if(x1 >= 0 && x1 < n && x2 >= 0 && x2 < n && y1 >= 0 && y1 < m && y2 >= 0 && y2 < m && !vis[opp][x2][y1] && (g[x1][y1] == 1 || g[x1][y1] == 2)&& (g[x2][y2] == 1 || g[x2][y2] == 2)){q.push(node(2, x2, y1, t.dis + 1));vis[opp][x2][y1] = t.dis + 1;}x1 = t.x + 1, x2 = t.x + 2, y1 = t.y, y2 = t.y, opp = 2;if(x1 >= 0 && x1 < n && x2 >= 0 && x2 < n && y1 >= 0 && y1 < m && y2 >= 0 && y2 < m && !vis[opp][x1][y1]&& (g[x1][y1] == 1 || g[x1][y1] == 2)&& (g[x2][y2] == 1 || g[x2][y2] == 2)){q.push(node(2, x1, y1, t.dis + 1));vis[opp][x1][y1] = t.dis + 1;}x1 = t.x, x2 = t.x, y1 = t.y + 1, y2 = t.y + 2, opp = 1;if(x1 >= 0 && x1 < n && x2 >= 0 && x2 < n && y1 >= 0 && y1 < m && y2 >= 0 && y2 < m && !vis[opp][x1][y1] && (g[x1][y1] == 1 || g[x1][y1] == 2)&& (g[x2][y2] == 1 || g[x2][y2] == 2)){q.push(node(1, x1, y1, t.dis + 1));vis[opp][x1][y1] = t.dis + 1;}x1 = t.x, x2 = t.x, y1 = t.y - 1, y2 = t.y - 2, opp = 1;if(x1 >= 0 && x1 < n && x2 >= 0 && x2 < n && y1 >= 0 && y1 < m && y2 >= 0 && y2 < m && !vis[opp][x2][y2] && (g[x1][y1] == 1 || g[x1][y1] == 2)&& (g[x2][y2] == 1 || g[x2][y2] == 2)){q.push(node(1, x2, y2, t.dis + 1));vis[opp][x2][y2] = t.dis + 1;}}else if(op1 == 1){x1 = t.x - 1, x2 = t.x - 1, y1 = t.y, y2 = t.y + 1, opp = 1;if(x1 >= 0 && x1 < n && x2 >= 0 && x2 < n && y1 >= 0 && y1 < m && y2 >= 0 && y2 < m && !vis[opp][x1][y1] && (g[x1][y1] == 1 || g[x1][y1] == 2)&& (g[x2][y2] == 1 || g[x2][y2] == 2)){q.push(node(1, x1, y1, t.dis + 1));vis[opp][x1][y1] = t.dis + 1;}x1 = t.x + 1, x2 = t.x + 1, y1 = t.y, y2 = t.y + 1, opp = 1;if(x1 >= 0 && x1 < n && x2 >= 0 && x2 < n && y1 >= 0 && y1 < m && y2 >= 0 && y2 < m && !vis[opp][x1][y1] && (g[x1][y1] == 1 || g[x1][y1] == 2)&& (g[x2][y2] == 1 || g[x2][y2] == 2)){q.push(node(1, x1, y1, t.dis + 1));vis[opp][x1][y1] = t.dis + 1;}x1 = t.x, x2 = 0, y1 = t.y - 1, y2 = 0, opp = 0;if(x1 >= 0 && x1 < n && x2 >= 0 && x2 < n && y1 >= 0 && y1 < m && y2 >= 0 && y2 < m && !vis[opp][x1][y1] && (g[x1][y1] == 1 || g[x1][y1] == 3)){q.push(node(0, x1, y1, t.dis + 1));vis[opp][x1][y1] = t.dis + 1;}x1 = t.x, x2 = 0, y1 = t.y + 2, y2 = 0, opp = 0;if(x1 >= 0 && x1 < n && x2 >= 0 && x2 < n && y1 >= 0 && y1 < m && y2 >= 0 && y2 < m && !vis[opp][x1][y1] && (g[x1][y1] == 1|| g[x1][y1] == 3)){q.push(node(0, x1, y1, t.dis + 1));vis[opp][x1][y1] = t.dis + 1;}}else if(op1 == 2){x1 = t.x , x2 = t.x + 1, y1 = t.y - 1, y2 = t.y - 1, opp = 2;if(x1 >= 0 && x1 < n && x2 >= 0 && x2 < n && y1 >= 0 && y1 < m && y2 >= 0 && y2 < m && !vis[opp][x1][y1] && (g[x1][y1] == 1 || g[x1][y1] == 2)&& (g[x2][y2] == 1 || g[x2][y2] == 2)){q.push(node(2, x1, y1, t.dis + 1));vis[opp][x1][y1] = t.dis + 1;}x1 = t.x, x2 = t.x + 1, y1 = t.y + 1, y2 = t.y + 1, opp = 2;if(x1 >= 0 && x1 < n && x2 >= 0 && x2 < n && y1 >= 0 && y1 < m && y2 >= 0 && y2 < m && !vis[opp][x1][y1] && (g[x1][y1] == 1 || g[x1][y1] == 2)&& (g[x2][y2] == 1 || g[x2][y2] == 2)){q.push(node(2, x1, y1, t.dis + 1));vis[opp][x1][y1] = t.dis + 1;}x1 = t.x - 1, x2 = 0, y1 = t.y, y2 = 0, opp = 0;if(x1 >= 0 && x1 < n && x2 >= 0 && x2 < n && y1 >= 0 && y1 < m && y2 >= 0 && y2 < m && !vis[opp][x1][y1] && (g[x1][y1] == 1|| g[x1][y1] == 3)){q.push(node(0, x1, y1, t.dis + 1));vis[opp][x1][y1] = t.dis + 1;}x1 = t.x + 2, x2 = 0, y1 = t.y, y2 = 0, opp = 0;if(x1 >= 0 && x1 < n && x2 >= 0 && x2 < n && y1 >= 0 && y1 < m && y2 >= 0 && y2 < m && !vis[opp][x1][y1] && (g[x1][y1] == 1|| g[x1][y1] == 3)){q.push(node(0, x1, y1, t.dis + 1));vis[opp][x1][y1] = t.dis + 1;}}}} int main(){cin>>n>>m;int q;for(int i = 0; i < n; i ++){for(int j = 0; j < m; j ++){scanf("%1d", &g[i][j]);if(g[i][j] == 3){ex = i, ey = j;}} }cin>>q;int x, y, op;bfs(0, ex, ey);while(q--){cin>>x>>y>>op;if(vis[op][x - 1][y - 1] || (op == 0 && x - 1 == ex && y - 1 == ey))cout<<vis[op][x - 1][y - 1]<<endl;else cout<<-1<<endl;}return 0; }