A - Airport Coffee
留坑。
B - Best Relay Team
枚举首棒
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 #define N 510 6 #define INF 0x3f3f3f3f 7 8 struct node 9 { 10 string name; 11 double v1, v2; 12 inline void scan() 13 { 14 cin >> name >> v1 >> v2; 15 } 16 inline bool operator < (const node &r) const 17 { 18 return v2 < r.v2; 19 } 20 }arr[N]; 21 22 int n; 23 double Min; 24 string ans[4], tmpname[4]; 25 26 int main() 27 { 28 cin.tie(0); 29 cout.tie(0); 30 ios::sync_with_stdio(false); 31 cin >> n; 32 for (int i = 1; i <= n; ++i) arr[i].scan(); 33 sort(arr + 1, arr + 1 + n); 34 Min = INF * 1.0; 35 for (int i = 1; i <= n; ++i) 36 { 37 double tmp = 0; 38 tmp += arr[i].v1; 39 tmpname[0] = arr[i].name; 40 for (int j = 1, cnt = 1; j <= n && cnt < 4; ++j) 41 { 42 if (i == j) continue; 43 tmp += arr[j].v2; 44 tmpname[cnt++] = arr[j].name; 45 } 46 if (tmp < Min) 47 { 48 Min = tmp; 49 for (int j = 0; j < 4; ++j) ans[j] = tmpname[j]; 50 } 51 } 52 cout << setiosflags(ios::fixed) << setprecision(2) << Min << endl; 53 for (int i = 0; i < 4; ++i) cout << ans[i] << endl; 54 return 0; 55 }
C - Compass Card Sales
按题意模拟即可
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 100010 5 6 int n; 7 set <int> se; 8 9 struct node 10 { 11 int pos, v, pre, nx; 12 inline node() {} 13 inline node(int _pos, int _v) 14 { 15 pos = _pos; 16 v = _v; 17 pre = -1, nx = -1; 18 } 19 inline bool operator < (const node &r) const 20 { 21 return v < r.v; 22 } 23 }R[N], G[N], B[N]; 24 25 struct DT 26 { 27 int r, g, b, id; 28 int id_r, id_g, id_b; 29 int v_r, v_g, v_b, v; 30 inline void scan() 31 { 32 scanf("%d%d%d%d", &r, &g, &b, &id); 33 v = 0; 34 } 35 inline void Get() 36 { 37 int pre, nx; 38 v_r = 0; pre = R[id_r].pre, nx = R[id_r].nx; 39 if (R[id_r].v != R[pre].v && R[id_r].v != R[nx].v) 40 { 41 v_r += (R[id_r].v - R[pre].v + 360) % 360; 42 v_r += (R[nx].v - R[id_r].v+ 360) % 360; 43 } 44 45 v_b = 0; pre = B[id_b].pre, nx = B[id_b].nx; 46 if (B[id_b].v != B[pre].v && B[id_b].v != B[nx].v) 47 { 48 v_b += (B[id_b].v - B[pre].v + 360) % 360; 49 v_b += (B[nx].v - B[id_b].v + 360) % 360; 50 } 51 52 v_g = 0; pre = G[id_g].pre, nx = G[id_g].nx; 53 if (G[id_g].v != G[pre].v && G[id_g].v != G[nx].v) 54 { 55 v_g += (G[id_g].v - G[pre].v + 360) % 360; 56 v_g += (G[nx].v - G[id_g].v + 360) % 360; 57 } 58 v = v_r + v_g + v_b; 59 } 60 inline void Remove() 61 { 62 int pre, nx; 63 pre = R[id_r].pre, nx = R[id_r].nx; 64 R[pre].nx = nx; R[nx].pre = pre; 65 se.insert(R[pre].pos); se.insert(R[nx].pos); 66 67 pre = G[id_g].pre, nx = G[id_g].nx; 68 G[pre].nx = nx; G[nx].pre = pre; 69 se.insert(G[pre].pos); se.insert(G[nx].pos); 70 71 pre = B[id_b].pre, nx = B[id_b].nx; 72 B[pre].nx = nx; B[nx].pre = pre; 73 se.insert(B[pre].pos); se.insert(B[nx].pos); 74 } 75 }arr[N]; 76 77 struct Node 78 { 79 int id, v, pos; 80 inline Node() {} 81 inline Node(int pos, int id, int v) : pos(pos), id(id), v(v) {} 82 inline bool operator < (const Node &r) const 83 { 84 return v > r.v || v == r.v && id < r.id; 85 } 86 }; 87 88 bool used[N]; 89 int sum[N]; 90 priority_queue <Node> q; 91 92 inline void Run() 93 { 94 while (scanf("%d", &n) != EOF) 95 { 96 while (!q.empty()) q.pop(); 97 memset(used, false, sizeof used); 98 memset(sum, 0, sizeof sum); 99 for (int i = 1; i <= n; ++i) 100 { 101 arr[i].scan(); 102 R[i] = node(i, arr[i].r); 103 G[i] = node(i, arr[i].g); 104 B[i] = node(i, arr[i].b); 105 } 106 if (n == 1) 107 { 108 printf("%d ", arr[1].id); 109 continue; 110 } 111 sort(R + 1, R + 1 + n); 112 sort(G + 1, G + 1 + n); 113 sort(B + 1, B + 1 + n); 114 for (int i = 1; i <= n; ++i) 115 { 116 arr[R[i].pos].id_r = i; 117 arr[G[i].pos].id_g = i; 118 arr[B[i].pos].id_b = i; 119 if (i == 1) 120 { 121 R[i].pre = n; G[i].pre = n; B[i].pre = n; 122 R[i].nx = i + 1; G[i].nx = i + 1; B[i].nx = i + 1; 123 } 124 else if (i == n) 125 { 126 R[i].pre = i - 1; G[i].pre = i - 1; B[i].pre = i - 1; 127 R[i].nx = 1; G[i].nx = 1; B[i].nx = 1; 128 } 129 else 130 { 131 R[i].pre = i - 1; G[i].pre = i - 1; B[i].pre = i - 1; 132 R[i].nx = i + 1; G[i].nx = i + 1; B[i].nx = i + 1; 133 } 134 } 135 for (int i = 1; i <= n; ++i) 136 { 137 arr[i].Get(); sum[i] = arr[i].v; 138 q.emplace(i, arr[i].id, sum[i]); 139 } 140 int cnt = 0; 141 while (cnt < n) 142 { 143 Node tmp = q.top(); q.pop(); 144 if (used[tmp.pos] || sum[tmp.pos] != tmp.v) continue; 145 ++cnt; printf("%d ", tmp.id); used[tmp.pos] = true; 146 se.clear(); arr[tmp.pos].Remove(); 147 for (auto it : se) 148 { 149 arr[it].Get(); 150 sum[it] = arr[it].v; 151 q.emplace(it, arr[it].id, sum[it]); 152 } 153 } 154 } 155 } 156 157 int main() 158 { 159 #ifdef LOCAL 160 freopen("Test.in", "r", stdin); 161 #endif 162 163 Run(); 164 165 return 0; 166 }
D - Distinctive Character
题意：给出n个玩家，有k个特征，用01串表示每个玩家有哪些特征，构造一个01串，和每个玩家对比有一个分数，有一个相同特征得一分，使得最大的分数最小
思路：可以将题意理解为有一个不同得一分，使得最小的分数最大。那么这时候可以用BFS跑一遍（类最短路？），然后找最大即可。
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 const int maxn = 1 << 21; 6 7 int n, k; 8 char str[100]; 9 int vis[maxn]; 10 int dis[maxn]; 11 12 inline void work(int x) 13 { 14 for(int i = k - 1; i >= 0; i--) 15 { 16 int u = 1 << i; 17 if(u & x) cout << "1"; 18 else cout << "0"; 19 } 20 cout << endl; 21 } 22 23 int main() 24 { 25 while(~scanf("%d %d", &n, &k)) 26 { 27 memset(vis, 0, sizeof vis); 28 memset(dis, -1, sizeof dis); 29 queue<int>q; 30 int tot = 1 << k; 31 for(int i = 1; i <= n; ++i) 32 { 33 scanf("%s", str); 34 int tmp = 0; 35 for(int i = 0; i < k; ++i) 36 { 37 tmp = tmp * 2 + str[i] - '0'; 38 } 39 vis[tmp] = 1; 40 } 41 for(int i = 0; i < tot; ++i) 42 { 43 if(vis[i] == 1) 44 { 45 q.push(i); 46 dis[i] = 0; 47 } 48 } 49 while(!q.empty()) 50 { 51 int st = q.front(); 52 q.pop(); 53 for(int i = 0 ; i < k; ++i) 54 { 55 int now = st ^ (1 << i); 56 if(dis[now] == -1) 57 { 58 // work(now); 59 dis[now] = dis[st] + 1; 60 q.push(now); 61 } 62 } 63 } 64 int Max = -1; 65 int ans = -1; 66 for(int i = 0; i < tot; ++i) 67 { 68 if(dis[i] > Max) 69 { 70 Max = dis[i]; 71 ans = i; 72 } 73 } 74 for(int i = k - 1; i >= 0; --i) 75 { 76 int u = 1 << i; 77 if(u & ans) 78 { 79 printf("1"); 80 } 81 else 82 { 83 printf("0"); 84 } 85 } 86 printf(" "); 87 } 88 return 0; 89 }
E - Emptying the Baltic
细胞(油田)。变种
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 typedef long long ll; 6 7 #define N 510 8 9 int n, m; 10 ll G[N][N]; 11 ll val[N][N]; 12 bool vis[N][N]; 13 int dir[][2] = {1,0,0,1,-1,0,0,-1,1,1,1,-1,-1,-1,-1,1}; 14 15 struct node{ 16 int x, y; 17 ll v; 18 inline node(){} 19 inline node(int x, int y, ll v) :x(x), y(y), v(v){} 20 inline bool operator < (const node & b)const 21 { 22 return v > b.v; 23 } 24 }; 25 26 inline bool judge(int x, int y) 27 { 28 if(x > n || x < 1 || y < 1 || y > m || vis[x][y] || G[x][y] >= 0) return false; 29 else return true; 30 } 31 32 int x, y; 33 34 inline void BFS() 35 { 36 memset(val, 0, sizeof val); 37 memset(vis, false, sizeof vis); 38 priority_queue<node>q; 39 q.push(node(x, y, G[x][y])); 40 vis[x][y] = true; 41 val[x][y] = G[x][y]; 42 while(!q.empty()) 43 { 44 node st = q.top(); 45 q.pop(); 46 for(int i = 0 ; i < 8; ++i) 47 { 48 int dx = st.x + dir[i][0]; 49 int dy = st.y + dir[i][1]; 50 if(judge(dx, dy)) 51 { 52 ll tmp = max(G[dx][dy], st.v); 53 val[dx][dy] = tmp; 54 G[dx][dy] = tmp; 55 q.push(node(dx, dy, tmp)); 56 vis[dx][dy] = true; 57 } 58 } 59 } 60 } 61 62 int main() 63 { 64 while(~scanf("%d %d", &n, &m)) 65 { 66 for(int i = 1; i <= n; ++i) 67 { 68 for(int j = 1; j <= m; ++j) 69 { 70 scanf("%lld", &G[i][j]); 71 if(G[i][j] >= 0) 72 { 73 G[i][j] = 0; 74 } 75 } 76 } 77 scanf("%d %d", &x, &y); 78 BFS(); 79 ll ans = 0; 80 for(int i = 1; i <= n; ++i) 81 { 82 for(int j = 1; j <= m; ++j) 83 { 84 if(val[i][j] < 0) 85 { 86 ans += -val[i][j]; 87 } 88 } 89 } 90 printf("%lld ", ans); 91 } 92 return 0; 93 }
F - Fractal Tree
留坑。
G - Galactic Collegiate Programming Contest
题意：给出ICPC过题记录，给出id为1的队伍的排名
思路：先hash, 离散化，数据结构维护一下
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 #define N 100010 6 #define ll long long 7 8 const ll D = (ll)1e9; 9 10 int n, m, q; 11 ll arr[N], brr[N], sum[N], cnt; 12 13 struct DT 14 { 15 int id; 16 ll v; 17 inline void scan() 18 { 19 scanf("%d%lld", &id, &v); 20 v = sum[id] + D - v; 21 sum[id] = v; 22 arr[++cnt] = v; 23 } 24 }Data[N]; 25 26 inline void Init_Hash() 27 { 28 for (int i = 1; i <= cnt; ++i) brr[i] = arr[i]; 29 sort(brr + 1, brr + 1 + cnt); 30 m = unique(brr + 1, brr + 1 + cnt) - brr - 1; 31 } 32 33 inline int Get(ll x) 34 { 35 return lower_bound(brr + 1, brr + 1 + m, x) - brr; 36 } 37 38 struct node 39 { 40 int l, r; 41 int sum; 42 inline node() {} 43 inline node(int l, int r, int sum) : l(l), r(r), sum(sum) {} 44 }tree[N << 2]; 45 46 inline void pushup(int id) 47 { 48 tree[id].sum = tree[id << 1].sum + tree[id << 1 | 1].sum; 49 } 50 51 inline void build(int id, int l, int r) 52 { 53 tree[id] = node(l, r, 0); 54 if (l == r) return; 55 int mid = (l + r) >> 1; 56 build(id << 1, l, mid); 57 build(id << 1 | 1, mid + 1, r); 58 } 59 60 inline void update(int id, int pos, int val) 61 { 62 if (tree[id].l == tree[id].r) 63 { 64 tree[id].sum += val; 65 return; 66 } 67 int mid = (tree[id].l + tree[id].r) >> 1; 68 if (pos <= mid) update(id << 1, pos, val); 69 else update(id << 1 | 1, pos, val); 70 pushup(id); 71 } 72 73 int anssum; 74 75 inline void query(int id, int l, int r) 76 { 77 if (l > r) return; 78 if (tree[id].l >= l && tree[id].r <= r) 79 { 80 anssum += tree[id].sum; 81 return; 82 } 83 int mid = (tree[id].l + tree[id].r) >> 1; 84 if (l <= mid) query(id << 1, l, r); 85 if (r > mid) query(id << 1 | 1, l, r); 86 } 87 88 int main() 89 { 90 while (scanf("%d%d", &n, &q) != EOF) 91 { 92 int id; ll v; 93 memset(sum, 0, sizeof sum); cnt = 0; 94 for (int i = 1; i <= q; ++i) Data[i].scan(); 95 Init_Hash(); 96 memset(sum, 0, sizeof sum); build(1, 1, q); 97 for (int i = 1; i <= q; ++i) 98 { 99 int id = Data[i].id; int pos = Get(Data[i].v); 100 if (sum[id]) update(1, sum[id], -1); 101 update(1, pos, 1); sum[id] = pos; 102 anssum = 0; query(1, sum[1] + 1, q); 103 printf("%d ", anssum + 1); 104 } 105 } 106 return 0; 107 }
H - Hubtown
留坑。
I - Import Spaghetti
题意：给出依赖关系，求最小的循环长度
思路：存图，枚举起点跑最短路
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 #define N 100010 6 #define INF 0x3f3f3f3f 7 8 unordered_map <string, int> mp; 9 int cnt, n; 10 string str[N], s; 11 vector <string> ans; 12 int Max, tot; 13 14 inline int Get(string s) 15 { 16 if (mp.find(s) == mp.end()) mp[s] = ++cnt, str[cnt] = s; 17 return mp[s]; 18 } 19 20 struct Edge 21 { 22 int to, nx; 23 inline Edge() {} 24 inline Edge(int to, int nx) : to(to), nx(nx) {} 25 }edge[N << 2]; 26 27 int head[N], pos; 28 29 inline void Init() 30 { 31 memset(head, -1, sizeof head); 32 pos = 0; cnt =0; mp.clear(); ans.clear(); Max = INF; 33 } 34 35 inline void addedge(int u, int v) 36 { 37 edge[++pos] = Edge(v, head[u]); head[u] = pos; 38 } 39 40 int dist[N]; 41 bool used[N]; 42 int pre[N]; 43 44 struct node 45 { 46 int to, w; 47 inline node() {} 48 inline node(int to, int w) : to(to), w(w) {} 49 inline bool operator < (const node &r) const 50 { 51 return w > r.w; 52 } 53 }; 54 55 inline void Dijkstra(int root) 56 { 57 for (int i = 1; i <= n; ++i) dist[i] = INF, used[i] = false, pre[i] = -1; 58 priority_queue <node> q; q.emplace(root, 0); 59 while (!q.empty()) 60 { 61 int u = q.top().to, w = q.top().w; q.pop(); 62 used[u] = true; 63 for (int it = head[u]; ~it; it = edge[it].nx) 64 { 65 int v = edge[it].to; 66 if (w + 1 < dist[v]) 67 { 68 pre[v] = u; 69 dist[v] = w + 1; 70 q.emplace(v, dist[v]); 71 } 72 } 73 } 74 } 75 76 int main() 77 { 78 cin.tie(0); cout.tie(0); 79 ios::sync_with_stdio(false); 80 while (cin >> n) 81 { 82 Init(); 83 // cout << "bug "; 84 for (int i = 1; i <= n; ++i) 85 { 86 cin >> s; Get(s); 87 // cout << s << endl; 88 } 89 // cout << "bug "; 90 for (int i = 1; i <= n; ++i) 91 { 92 cin >> s >> tot; 93 int u = Get(s); 94 getline(cin, s); 95 while (tot--) 96 { 97 getline(cin, s); 98 // cout << s << endl; 99 stringstream ss; 100 ss << s; 101 while (ss >> s) 102 { 103 if (s == "import") continue; 104 if (s.end()[-1] == ',') s.erase(s.end() - 1); 105 addedge(u, Get(s)); 106 } 107 } 108 } 109 for (int i = 1; i <= n; ++i) 110 { 111 Dijkstra(i); 112 if (dist[i] < Max) 113 { 114 Max = dist[i]; 115 int it = pre[i]; 116 ans.clear(); 117 while (it != i) 118 { 119 ans.emplace_back(str[it]); 120 it = pre[it]; 121 } 122 ans.emplace_back(str[i]); 123 } 124 } 125 if (Max == INF) 126 { 127 cout << "SHIP IT" << endl; 128 continue; 129 } 130 reverse(ans.begin(), ans.end()); 131 for (int i = 0, len = ans.size(); i < len; ++i) cout << ans[i] << " "[i == len -1]; 132 } 133 return 0; 134 }
J - Judging Moose
签到
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 int a, b; 6 7 int main() 8 { 9 while (scanf("%d%d", &a, &b) != EOF) 10 { 11 if (a == 0 && b == 0) 12 puts("Not a moose"); 13 else if (a == b) 14 { 15 printf("Even %d ", a + b); 16 } 17 else 18 printf("Odd %d ", max(a, b) * 2); 19 } 20 return 0; 21 }
K - Kayaking Trip
题意：给出三种人，每种人有力量值，然后有(人数总和)/2 条皮划艇，一条皮划艇的速度为 $v = c_i(s_1 + s_2)$, 求如何分配人员，使得速度最小的皮划艇速度最大
思路：二分答案，贪心check
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 #define N 100010 6 #define INF 0x3f3f3f3f 7 8 int n[3], s[3], m, tmp[3]; 9 int arr[N]; 10 11 struct node{ 12 int x, y, v; 13 inline node() {} 14 inline node(int x, int y, int v) : x(x), y(y), v(v) {} 15 inline bool operator < (const node &b) const 16 { 17 return v < b.v; 18 } 19 }brr[10]; 20 21 int cnt; 22 23 inline bool check(int mid) 24 { 25 for (int i = 0; i < 3; ++i) tmp[i] = n[i]; 26 for (int i = 1; i <= m; ++i) 27 { 28 bool flag = false; 29 for (int j = 1; j <= cnt; ++j) 30 { 31 int x = brr[j].x, y = brr[j].y; 32 if (arr[i] * brr[j].v >= mid) 33 { 34 if (x == y) 35 { 36 if (tmp[x] >= 2) 37 { 38 tmp[x] -= 2; 39 flag = true; 40 break; 41 } 42 } 43 else 44 { 45 if (tmp[x] >= 1 && tmp[y] >= 1) 46 { 47 --tmp[x], --tmp[y]; 48 flag = true; 49 break; 50 } 51 } 52 } 53 } 54 if (flag == false) return false; 55 } 56 return true; 57 } 58 59 int main() 60 { 61 while (scanf("%d%d%d", &n[0], &n[1], &n[2]) != EOF) 62 { 63 m = n[0] + n[1] + n[2]; 64 m >>= 1; 65 for (int i = 0; i < 3; ++i) scanf("%d", s + i); 66 for (int i = 1; i <= m; ++i) scanf("%d", arr + i); 67 cnt = 0; 68 for (int i = 0; i < 3; ++i) 69 for (int j = i; j < 3; ++j) 70 brr[++cnt] = node(i, j, s[i] + s[j]); 71 sort(brr + 1, brr + 1 + cnt); 72 int l = 0, r = INF, ans; 73 while (r - l >= 0) 74 { 75 int mid = (l + r) >> 1; 76 if (check(mid)) 77 { 78 ans = mid; 79 l = mid + 1; 80 } 81 else 82 { 83 r = mid - 1; 84 } 85 } 86 printf("%d ", ans); 87 } 88 return 0; 89 }