PAT 甲级真题题解（121-155）
1121 Damn Single
模拟
1 // 1121 Damn Single 2 #include <map> 3 #include <vector> 4 #include <cstdio> 5 #include <iostream> 6 #include <algorithm> 7 using namespace std; 8 9 map<int, int> m, vis; 10 vector<int> p; 11 const int N = 1e4 + 10; 12 int a[N]; 13 14 int main() { 15 int n, x, y; 16 scanf("%d", &n); 17 while (n--) { 18 scanf("%d %d", &x, &y); 19 x++; y++; 20 m[x] = y; 21 m[y] = x; 22 } 23 scanf("%d", &n); 24 for (int i = 1; i <= n; i++) { 25 scanf("%d", &a[i]); 26 a[i]++; 27 vis[a[i]] = 1; 28 } 29 for (int i = 1; i <= n; i++) { 30 if (vis[ m[a[i]] ]) continue; 31 p.push_back(a[i]); 32 } 33 sort(p.begin(), p.end()); 34 printf("%d\n", p.size()); 35 for (int i = 0; i < p.size(); i++) { 36 if (i != 0) printf(" "); 37 printf("%05d", p[i] - 1); 38 } 39 return 0; 40 }
1122 Hamiltonian Cycle
模拟
1 // 1122 Hamiltonian Cycle 2 #include <set> 3 #include <cstdio> 4 #include <cstring> 5 #include <iostream> 6 #include <algorithm> 7 using namespace std; 8 9 const int N = 205; 10 int MAP[N][N]; 11 set<int> s; 12 13 int main() { 14 memset(MAP, -1, sizeof(MAP)); 15 int n, m, k; 16 scanf("%d %d", &n, &m); 17 for (int i = 1; i <= m; i++) { 18 int u, v; 19 scanf("%d %d", &u, &v); 20 MAP[u][v] = 1; MAP[v][u] = 1; 21 } 22 scanf("%d", &k); 23 while (k--) { 24 bool f = 1; 25 int u, v, root; 26 scanf("%d", &m); 27 scanf("%d", &root); 28 s.insert(root); 29 u = root; 30 for (int i = 1; i < m; i++) { 31 scanf("%d", &v); 32 s.insert(v); 33 if (MAP[u][v] == -1) f = 0; 34 u = v; 35 } 36 if (root != u || s.size() != n || m != n + 1) f = 0; 37 if (!f) printf("NO\n"); 38 else printf("YES\n"); 39 s.clear(); 40 } 41 return 0; 42 }
1124 Raffle for Weibo Followers
MAP标记
1 // 1124 Raffle for Weibo Followers 2 #include <map> 3 #include <cstdio> 4 #include <vector> 5 #include <iostream> 6 #include <algorithm> 7 using namespace std; 8 9 map<string, int> vis; 10 string str; 11 12 int main() { 13 bool f = 0; 14 int n, m, s, cnt; 15 cin >> n >> m >> s; 16 cnt = m; 17 for (int i = 1; i < s; i++) cin >> str; 18 for (int i = s; i <= n; i++) { 19 cin >> str; 20 if (cnt == m && !vis[str]) { 21 f = 1; 22 cout << str << endl; 23 cnt = 1; 24 vis[str] = 1; 25 } 26 if (!vis[str]) cnt++; 27 } 28 if (!f) cout << "Keep going..." << endl; 29 return 0; 30 }
1125 Chain the Ropes
思维
1 // 1125 Chain the Ropes 2 #include <cstdio> 3 #include <iostream> 4 #include <algorithm> 5 using namespace std; 6 7 const int N = 1e4 + 10; 8 int a[N]; 9 10 int main() { 11 int n, ans = 0; 12 cin >> n; 13 for (int i = 1; i <= n; i++) cin >> a[i]; 14 sort(a + 1, a + 1 + n); 15 ans = a[1]; 16 for (int i = 2; i <= n; i++) { 17 ans += a[i]; 18 ans /= 2; 19 } 20 cout << ans << endl; 21 return 0; 22 }
1126 Eulerian Path
给定m条边关系，判断是欧拉回路还是半欧拉回路或者不是欧拉回路。
如果一个连通图，具有0个奇度顶点为欧拉回路，具有2个奇度顶点为半欧拉，其他均不为欧拉图。
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 const int N = 1e5 + 5; 5 int cnt[N], fa[N]; 6 7 void init() { 8 for (int i = 0; i < N; i++) fa[i] = i; 9 } 10 11 int fi(int x) { 12 return fa[x] == x ? fa[x] : fa[x] = fi(fa[x]); 13 } 14 15 void Union(int x, int y) { 16 int fx = fi(x), fy = fi(y); 17 if (fx != fy) { 18 fa[fx] = fy; 19 } 20 } 21 22 int main() { 23 init(); 24 int n, m, ans = 0, s = 0; 25 scanf("%d %d", &n, &m); 26 for (int i = 1; i <= m; i++) { 27 int u, v; 28 scanf("%d %d", &u, &v); 29 Union(u, v); 30 cnt[u]++; cnt[v]++; 31 } 32 for (int i = 1; i <= n; i++) { 33 if (cnt[i] % 2) ans++; 34 if (fa[i] == i) s++; 35 if (i != 1) printf(" "); 36 printf("%d", cnt[i]); 37 } 38 printf("\n"); 39 if (s == 1 && ans <= 2) { 40 if (ans == 0) printf("Eulerian\n"); 41 else if (ans == 2) printf("Semi-Eulerian\n"); 42 else printf("Non-Eulerian\n"); 43 } else printf("Non-Eulerian\n"); 44 return 0; 45 }
1127 ZigZagging on a Tree
中序后序建树，层次遍历。（参考了柳神的做法，简洁多了。）
1 #include <queue> 2 #include <vector> 3 #include <iostream> 4 using namespace std; 5 6 vector<int> in, post, result[35]; 7 int n, tree[35][2], root; 8 9 struct node { 10 int index, depth; 11 }; 12 13 void dfs(int &index, int inLeft, int inRight, int postLeft, int postRight) { 14 if (inLeft > inRight) return; 15 index = postRight; 16 int i = 0; 17 while (in[i] != post[postRight]) i++; 18 dfs(tree[index][0], inLeft, i - 1, postLeft, postLeft + (i - inLeft) - 1); 19 dfs(tree[index][1], i + 1, inRight, postLeft + (i - inLeft), postRight - 1); 20 } 21 22 void bfs() { 23 queue<node> q; 24 q.push(node{root, 0}); 25 while (!q.empty()) { 26 node temp = q.front(); 27 q.pop(); 28 result[temp.depth].push_back(post[temp.index]); 29 if (tree[temp.index][0] != 0) 30 q.push(node{tree[temp.index][0], temp.depth + 1}); 31 if (tree[temp.index][1] != 0) 32 q.push(node{tree[temp.index][1], temp.depth + 1}); 33 } 34 } 35 36 int main() { 37 cin >> n; 38 in.resize(n + 1), post.resize(n + 1); 39 for (int i = 1; i <= n; i++) cin >> in[i]; 40 for (int i = 1; i <= n; i++) cin >> post[i]; 41 dfs(root, 1, n, 1, n); 42 bfs(); 43 printf("%d", result[0][0]); 44 for (int i = 1; i < 35; i++) { 45 if (i % 2 == 1) { 46 for (int j = 0; j < result[i].size(); j++) 47 printf(" %d", result[i][j]); 48 } else { 49 for (int j = result[i].size() - 1; j >= 0; j--) 50 printf(" %d", result[i][j]); 51 } 52 } 53 return 0; 54 }
1128 N Queens Puzzle
同行，同列，同斜判断下即可。
1 // 1128 N Queens Puzzle 2 #include <cstdio> 3 #include <iostream> 4 #include <algorithm> 5 using namespace std; 6 7 const int N = 1e5 + 10; 8 bool vis1[N], vis2[N]; 9 10 int main() { 11 int t, n, x; 12 cin >> t; 13 while (t--) { 14 bool ok = 1; 15 cin >> n; 16 for (int i = 0; i <= 3 * n; i++) vis1[i] = vis2[i] = 0; 17 for (int i = 1; i <= n; i++) { 18 cin >> x; 19 if (vis1[x] || vis2[x + i]) { 20 ok = 0; 21 } 22 vis1[x] = 1; 23 vis2[x + i] = 1; 24 } 25 if (ok) cout << "YES" << endl; 26 else cout << "NO" << endl; 27 } 28 return 0; 29 }
1129 Recommendation System
STL set应用
1 // 1129 Recommendation System 2 #include <set> 3 #include <cstdio> 4 #include <iostream> 5 #include <algorithm> 6 using namespace std; 7 8 const int N = 5e4 + 10; 9 struct node { 10 int id, time; 11 friend bool operator < (node x,node y){ 12 if(x.time == y.time) return x.id < y.id; 13 return x.time > y.time; 14 } 15 }; 16 17 int cnt[N]; 18 set<node> se; 19 set<node> ::iterator it; 20 21 int main() { 22 int n, k, x; 23 cin >> n >> k; 24 for (int i = 1; i <= n; i++) { 25 cin >> x; 26 if(i == 1) { 27 cnt[x]++; 28 se.insert({x, cnt[x]}); 29 continue; 30 } 31 else { 32 cout << x << ":"; 33 int c = 0; 34 for (auto y : se) { 35 c++; 36 if(c > k) break; 37 cout << " " << y.id; 38 } 39 cout << endl; 40 } 41 if (se.find({x, cnt[x]}) != se.end()) { 42 it = se.find({x, cnt[x]}); 43 se.erase(it); 44 } 45 cnt[x]++; 46 se.insert({x, cnt[x]}); 47 } 48 return 0; 49 }
1130 Infix Expression
给定中缀表达式二叉树，输出中缀表达式。（记得CSP也考过，当时是直接暴力A的。）
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 21 5 int n, Root; 6 int vis[N]; 7 8 struct Node { 9 string s; 10 int l, r; 11 } node[N]; 12 13 14 void dfs(int root) { 15 if (root == -1) return ; 16 if (root != Root && (node[root].l != -1 || node[root].r != -1)) cout << "("; 17 dfs(node[root].l); 18 cout << node[root].s; 19 dfs(node[root].r); 20 if ( root != Root && (node[root].l != -1 || node[root].r != -1)) cout<<")"; 21 } 22 23 int main() { 24 scanf("%d", &n); 25 memset(vis, 0, sizeof(vis)); 26 for (int i = 1; i <= n; i++) { 27 cin >> node[i].s >> node[i].l >> node[i].r; 28 vis[node[i].l] = vis[node[i].r] = 1; 29 } 30 Root = 1; 31 while (vis[Root]) Root++; 32 dfs(Root); 33 return 0; 34 }
1131 Subway Map
DFS暴力找最短路。同时保证题目中的那些要求。
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define MAX 10005 5 #define INF 0x3f3f3f3f 6 7 int N, M, K; 8 int S, E; 9 10 struct Stop { 11 int line; 12 int id; 13 } stops[MAX]; 14 15 vector<Stop> stopVec[MAX]; 16 int flag[MAX]; 17 int cntMax = INF; 18 vector<Stop> stopAns, stopsVec; 19 int subMin = 0; 20 21 void dfs(int x, int cnt) { 22 if(x == E) { 23 int sub = 0; 24 if(cntMax > cnt) { 25 stopAns = stopsVec; 26 cntMax = cnt; 27 for(int i = 1; i < stopsVec.size(); i++) { 28 if(stopsVec[i].line != stopsVec[i-1].line) { 29 sub++; 30 } 31 } 32 subMin = sub; 33 } 34 if(cntMax == cnt) { 35 for(int i = 1; i < stopsVec.size(); i++) { 36 if(stopsVec[i].line != stopsVec[i-1].line) { 37 sub++; 38 } 39 } 40 if(sub < subMin) { 41 stopAns = stopsVec; 42 subMin = sub; 43 } 44 } 45 return; 46 } 47 for(int i = 0; i < stopVec[x].size(); i++) { 48 if(flag[stopVec[x][i].id] == 0) { 49 flag[stopVec[x][i].id] = 1; 50 stopsVec.push_back(stopVec[x][i]); 51 dfs(stopVec[x][i].id, cnt+1); 52 stopsVec.pop_back(); 53 flag[stopVec[x][i].id] = 0; 54 } 55 } 56 } 57 58 int main() 59 { 60 cin>>N; 61 for(int i = 1; i <= N; i++) { 62 scanf("%d", &M); 63 for(int j = 0; j < M; j++) { 64 scanf("%d", &stops[j].id); 65 stops[j].line = i; 66 if(j != 0) { 67 stopVec[stops[j].id].push_back(stops[j-1]); 68 stopVec[stops[j-1].id].push_back(stops[j]); 69 } 70 } 71 } 72 cin>>K; 73 for(int i = 0; i < K; i++) { 74 scanf("%d%d", &S, &E); 75 cntMax = INF; 76 dfs(S, 0); 77 printf("%d\n", cntMax); 78 int line = stopAns[0].line; 79 int id = S; 80 for(int i = 0; i < stopAns.size(); i++) { 81 if(line != stopAns[i].line) { 82 printf("Take Line#%d from %04d to %04d.\n", line, id, stopAns[i-1].id); 83 line = stopAns[i].line; 84 id = stopAns[i-1].id; 85 } 86 } 87 printf("Take Line#%d from %04d to %04d.\n", line, id, E); 88 } 89 return 0; 90 }
1132 Cut Integer
注意判断拆开后是否有0。
1 #include <iostream> 2 using namespace std; 3 4 int cal(string s) { 5 int res = 0; 6 for (int i = 0; i < s.size(); i++) { 7 res = res * 10 + (s[i] - '0'); 8 } 9 return res; 10 } 11 12 int main() { 13 int n; 14 cin >> n; 15 while (n--) { 16 string m; 17 int a, b, c; 18 cin >> m; 19 c = cal(m); 20 a = cal(m.substr(0, m.size() / 2)); 21 b = cal(m.substr(m.size() / 2, m.size() / 2)); 22 if (a * b != 0 && c % (a * b) == 0) cout << "Yes" << endl; 23 else cout << "No" << endl; 24 } 25 return 0; 26 }
1133 Splitting A Linked List
从root开始，保存小于0的，大于等于0小于等于K的和大于K的链表，最后按顺序输出结果即可。
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 const int INF = 0x3f3f3f3f; 5 6 struct Node { 7 int data, next; 8 } node[100005]; 9 10 int root, a, N, K; 11 vector<int> v[3]; 12 13 int main() { 14 scanf("%d%d%d", &root, &N, &K); 15 for (int i = 0; i < N; i++) { 16 scanf("%d", &a); 17 scanf("%d %d", &node[a].data, &node[a].next); 18 } 19 while (root != -1) { 20 if (node[root].data < 0) v[0].push_back(root); 21 else if (node[root].data <= K) v[1].push_back(root); 22 else v[2].push_back(root); 23 root = node[root].next; 24 } 25 int f = 0; 26 for (int i = 0; i < 3; i++) { 27 for (int j = 0; j < v[i].size(); j++) { 28 if (!f) printf("%05d %d",v[i][j],node[v[i][j]].data), f = 1; 29 else printf(" %05d\n%05d %d",v[i][j],v[i][j],node[v[i][j]].data), f = 1; 30 } 31 } 32 printf(" -1\n"); 33 return 0; 34 }
