PAT 甲级真题题解（121-155）

1121 Damn Single

模拟

 // 1121 Damn Single
 #include <map>
 #include <vector>
 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 map<int, int> m, vis;
 vector<int> p;
 const int N = 1e4 + ;
 int a[N];

 int main() {
     int n, x, y;
     scanf("%d", &n);
     while (n--) {
         scanf("%d %d", &x, &y);
         x++; y++;
         m[x] = y;
         m[y] = x;
     }
     scanf("%d", &n);
     for (int i = ; i <= n; i++) {
         scanf("%d", &a[i]);
         a[i]++;
         vis[a[i]] = ;
     }
     for (int i = ; i <= n; i++) {
         if (vis[ m[a[i]] ]) continue;
         p.push_back(a[i]);
     }
     sort(p.begin(), p.end());
     printf("%d\n", p.size());
     for (int i = ; i < p.size(); i++) {
         if (i != ) printf(" ");
         printf("%05d", p[i] - );
     }
     return ;
 }

1122 Hamiltonian Cycle

模拟

 // 1122 Hamiltonian Cycle
 #include <set>
 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 const int N = ;
 int MAP[N][N];
 set<int> s;

 int main() {
     memset(MAP, -, sizeof(MAP));
     int n, m, k;
     scanf("%d %d", &n, &m);
     for (int i = ; i <= m; i++) {
         int u, v;
         scanf("%d %d", &u, &v);
         MAP[u][v] = ; MAP[v][u] = ;
     }
     scanf("%d", &k);
     while (k--) {
         bool f = ;
         int u, v, root;
         scanf("%d", &m);
         scanf("%d", &root);
         s.insert(root);
         u = root;
         for (int i = ; i < m; i++) {
             scanf("%d", &v);
             s.insert(v);
             if (MAP[u][v] == -) f = ;
             u = v;
         }
         if (root != u || s.size() != n || m != n + ) f = ;
         if (!f) printf("NO\n");
         else printf("YES\n");
         s.clear();
     }
     return ;
 }

1124 Raffle for Weibo Followers

MAP标记

 // 1124 Raffle for Weibo Followers
 #include <map>
 #include <cstdio>
 #include <vector>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 map<string, int> vis;
 string str;

 int main() {
     bool f = ;
     int n, m, s, cnt;
     cin >> n >> m >> s;
     cnt = m;
     for (int i = ; i < s; i++) cin >> str;
     for (int i = s; i <= n; i++) {
         cin >> str;
         if (cnt == m && !vis[str]) {
             f = ;
             cout << str << endl;
             cnt = ;
             vis[str] = ;
         }
         if (!vis[str]) cnt++;
     }
     if (!f) cout << "Keep going..." << endl;
     return ;
 }

1125 Chain the Ropes

思维

 // 1125 Chain the Ropes
 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 const int N = 1e4 + ;
 int a[N];

 int main() {
     int n, ans = ;
     cin >> n;
     for (int i = ; i <= n; i++) cin >> a[i];
     sort(a + , a +  + n);
     ans = a[];
     for (int i = ; i <= n; i++) {
         ans += a[i];
         ans /= ;
     }
     cout << ans << endl;
     return ;
 }

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。

 #include <bits/stdc++.h>
 using namespace std;

 const int N = 1e5 + ;
 int cnt[N], fa[N];

 void init() {
     for (int i = ; i < N; i++) fa[i] = i;
 }

 int fi(int x) {
     return fa[x] == x ? fa[x] : fa[x] = fi(fa[x]);
 }

 void Union(int x, int y) {
     int fx = fi(x), fy = fi(y);
     if (fx != fy) {
         fa[fx] = fy;
     }
 }

 int main() {
     init();
     int n, m, ans = , s = ;
     scanf("%d %d", &n, &m);
     for (int i = ; i <= m; i++) {
         int u, v;
         scanf("%d %d", &u, &v);
         Union(u, v);
         cnt[u]++; cnt[v]++;
     }
     for (int i = ; i <= n; i++) {
         if (cnt[i] % ) ans++;
         if (fa[i] == i) s++;
         if (i != ) printf(" ");
         printf("%d", cnt[i]);
     }
     printf("\n");
     if (s ==  && ans <= ) {
         if (ans == ) printf("Eulerian\n");
         else if (ans == ) printf("Semi-Eulerian\n");
         else printf("Non-Eulerian\n");
     } else printf("Non-Eulerian\n");
     return ;
 }

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。）

 #include <queue>
 #include <vector>
 #include <iostream>
 using namespace std;

 vector<int> in, post, result[];
 int n, tree[][], root;

 struct node {
     int index, depth;
 };

 void dfs(int &index, int inLeft, int inRight, int postLeft, int postRight) {
     if (inLeft > inRight) return;
     index = postRight;
     int i = ;
     while (in[i] != post[postRight]) i++;
     dfs(tree[index][], inLeft, i - , postLeft, postLeft + (i - inLeft) - );
     dfs(tree[index][], i + , inRight, postLeft + (i - inLeft), postRight - );
 }

 void bfs() {
     queue<node> q;
     q.push(node{root, });
     while (!q.empty()) {
         node temp = q.front();
         q.pop();
         result[temp.depth].push_back(post[temp.index]);
         if (tree[temp.index][] != )
             q.push(node{tree[temp.index][], temp.depth + });
         if (tree[temp.index][] != )
             q.push(node{tree[temp.index][], temp.depth + });
     }
 }

 int main() {
     cin >> n;
     in.resize(n + ), post.resize(n + );
     for (int i = ; i <= n; i++) cin >> in[i];
     for (int i = ; i <= n; i++) cin >> post[i];
     dfs(root, , n, , n);
     bfs();
     printf("%d", result[][]);
     for (int i = ; i < ; i++) {
         if (i %  == ) {
             for (int j = ; j < result[i].size(); j++)
                 printf(" %d", result[i][j]);
         } else {
             for (int j = result[i].size() - ; j >= ; j--)
                 printf(" %d", result[i][j]);
         }
     }
     return ;
 }

1128 N Queens Puzzle

同行，同列，同斜判断下即可。

 // 1128 N Queens Puzzle
 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 const int N = 1e5 + ;
 bool vis1[N], vis2[N];

 int main() {
     int t, n, x;
     cin >> t;
     while (t--) {
         bool ok = ;
         cin >> n;
         for (int i = ; i <=  * n; i++) vis1[i] = vis2[i] = ;
         for (int i = ; i <= n; i++) {
             cin >> x;
             if (vis1[x] || vis2[x + i]) {
                 ok = ;
             }
             vis1[x] = ;
             vis2[x + i] = ;
         }
         if (ok) cout << "YES" << endl;
         else cout << "NO" << endl;
     }
     return ;
 }

1129 Recommendation System 

STL set应用

 // 1129 Recommendation System
 #include <set>
 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 const int N = 5e4 + ;
 struct node {
     int id, time;
     friend bool operator < (node x,node y){
         if(x.time == y.time) return x.id < y.id;
         return x.time > y.time;
     }
 };

 int cnt[N];
 set<node> se;
 set<node> ::iterator it;

 int main() {
     int n, k, x;
     cin >> n >> k;
     for (int i = ; i <= n; i++) {
         cin >> x;
         if(i == ) {
             cnt[x]++;
             se.insert({x, cnt[x]});
             continue;
         }
         else {
             cout << x << ":";
             int c = ;
             for (auto y : se) {
                 c++;
                 if(c > k) break;
                 cout << " " << y.id;
             }
             cout << endl;
         }
         if (se.find({x, cnt[x]}) != se.end()) {
             it = se.find({x, cnt[x]});
             se.erase(it);
         }
         cnt[x]++;
         se.insert({x, cnt[x]});
     }
     return ;
 }

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。）

 #include <bits/stdc++.h>
 using namespace std;

 #define N 21
 int n, Root;
 int vis[N];

 struct Node {
     string s;
     int l, r;
 } node[N];

 void dfs(int root) {
     if (root == -) return ;
     if (root != Root && (node[root].l != - || node[root].r != -)) cout << "(";
     dfs(node[root].l);
     cout << node[root].s;
     dfs(node[root].r);
     if ( root != Root && (node[root].l != - || node[root].r != -)) cout<<")";
 }

 int main() {
     scanf("%d", &n);
     memset(vis, , sizeof(vis));
     for (int i = ; i <= n; i++) {
         cin >> node[i].s >> node[i].l >> node[i].r;
         vis[node[i].l] = vis[node[i].r] = ;
     }
     Root = ;
     while (vis[Root]) Root++;
     dfs(Root);
     return ;
 }

1131 Subway Map

DFS暴力找最短路。同时保证题目中的那些要求。

 #include <bits/stdc++.h>
 using namespace std;

 #define MAX 10005
 #define INF 0x3f3f3f3f

 int N, M, K;
 int S, E;

 struct Stop {
     int line;
     int id;
 } stops[MAX];

 vector<Stop> stopVec[MAX];
 int flag[MAX];
 int cntMax = INF;
 vector<Stop> stopAns, stopsVec;
 int subMin = ;

 void dfs(int x, int cnt) {
     if(x == E) {
         int sub = ;
         if(cntMax > cnt) {
             stopAns = stopsVec;
             cntMax = cnt;
             for(int i = ; i < stopsVec.size(); i++) {
                 if(stopsVec[i].line != stopsVec[i-].line) {
                     sub++;
                 }
             }
             subMin = sub;
         }
         if(cntMax == cnt) {
             for(int i = ; i < stopsVec.size(); i++) {
                 if(stopsVec[i].line != stopsVec[i-].line) {
                     sub++;
                 }
             }
             if(sub < subMin) {
                 stopAns = stopsVec;
                 subMin = sub;
             }
         }
         return;
     }
     for(int i = ; i < stopVec[x].size(); i++) {
         if(flag[stopVec[x][i].id] == ) {
             flag[stopVec[x][i].id] = ;
             stopsVec.push_back(stopVec[x][i]);
             dfs(stopVec[x][i].id, cnt+);
             stopsVec.pop_back();
             flag[stopVec[x][i].id] = ;
         }
     }
 }

 int main()
 {
     cin>>N;
     for(int i = ; i <= N; i++) {
         scanf("%d", &M);
         for(int j = ; j < M; j++) {
             scanf("%d", &stops[j].id);
             stops[j].line = i;
             if(j != ) {
                 stopVec[stops[j].id].push_back(stops[j-]);
                 stopVec[stops[j-].id].push_back(stops[j]);
             }
         }
     }
     cin>>K;
     for(int i = ; i < K; i++) {
         scanf("%d%d", &S, &E);
         cntMax = INF;
         dfs(S, );
         printf("%d\n", cntMax);
         int line = stopAns[].line;
         int id = S;
         for(int i = ; i < stopAns.size(); i++) {
             if(line != stopAns[i].line) {
                 printf("Take Line#%d from %04d to %04d.\n", line, id, stopAns[i-].id);
                 line = stopAns[i].line;
                 id = stopAns[i-].id;
             }
         }
         printf("Take Line#%d from %04d to %04d.\n", line, id, E);
     }
     return ;
 }

1132 Cut Integer  

注意判断拆开后是否有0。

 #include <iostream>
 using namespace std;

 int cal(string s) {
     int res = ;
     for (int i = ; i < s.size(); i++) {
         res = res *  + (s[i] - '');
     }
     return res;
 }

 int main() {
     int n;
     cin >> n;
     while (n--) {
         string m;
         int a, b, c;
         cin >> m;
         c = cal(m);
         a = cal(m.substr(, m.size() / ));
         b = cal(m.substr(m.size() / , m.size() / ));
         if (a * b !=  && c % (a * b) == ) cout << "Yes" << endl;
         else cout << "No" << endl;
     }
     return ;
 }

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。

 #include <bits/stdc++.h>
 using namespace std;

 const int INF = 0x3f3f3f3f;

 struct Node {
     int data, next;
 } node[];

 int root, a, N, K;
 vector<int> v[];

 int main() {
     scanf("%d%d%d", &root, &N, &K);
     for (int i = ; i < N; i++) {
         scanf("%d", &a);
         scanf("%d %d", &node[a].data, &node[a].next);
     }
     while (root != -) {
         if (node[root].data < ) v[].push_back(root);
         else if (node[root].data <= K) v[].push_back(root);
         else v[].push_back(root);
         root = node[root].next;
     }
     int f = ;
     for (int i = ; i < ; i++) {
         for (int j = ; j < v[i].size(); j++) {
             if (!f) printf("%05d %d",v[i][j],node[v[i][j]].data), f = ;
             else printf(" %05d\n%05d %d",v[i][j],v[i][j],node[v[i][j]].data), f = ;
         }
     }
     printf(" -1\n");
     return ;
 }