There is a famous railway station in PopPush City. Country there is incredibly hilly. The station was built in last century. Unfortunately, funds were extremely limited that time. It was possible to establish only a surface track. Moreover, it turned out that the station could be only a dead-end one (see picture) and due to lack of available space it could have only one track.
The local tradition is that every train arriving from the direction A continues in the direction B with coaches reorganized in some way. Assume that the train arriving from the direction A has N ≤ 1000 coaches numbered in increasing order 1, 2, . . . , N. The chief for train reorganizations must know whether it is possible to marshal coaches continuing in the direction B so that their order will be a1.a2, . . . , aN . Help him and write a program that decides whether it is possible to get the required order of coaches. You can assume that single coaches can be disconnected from the train before they enter the station and that they can move themselves until they are on the track in the direction B. You can also suppose that at any time there can be located as many coaches as necessary in the station. But once a coach has entered the station it cannot return to the track in the direction A and also once it has left the station in the direction B it cannot return back to the station.
Input
The input file consists of blocks of lines. Each block except the last describes one train and possibly more requirements for its reorganization. In the first line of the block there is the integer N described above. In each of the next lines of the block there is a permutation of 1, 2, . . . , N. The last line of the block contains just ‘0’. The last block consists of just one line containing ‘0’.
Output
The output file contains the lines corresponding to the lines with permutations in the input file. A line of the output file contains ‘Yes’ if it is possible to marshal the coaches in the order required on the corresponding line of the input file. Otherwise it contains ‘No’. In addition, there is one empty line after the lines corresponding to one block of the input file. There is no line in the output file corresponding to the last “null” block of the input file.
Sample Input
5
1 2 3 4 5
5 4 1 2 3
0
6
6 5 4 3 2 1
0
0
Sample Output
Yes
No
Yes
题目类型：栈模拟
算法分析：使用一个Stack数组表示序列中数的情况，Stack[i] == 0表示i元素还没有入栈；Stack[i] == 1表示i元素在栈内；Stack[i] == 2表示i元素已经出栈。边输入边处理：维护一个合法标志valid，初始时valid置为true。读入一个元素a，判断比a大的元素是否还有在栈中的，如果还有则valid置为false；如果判断完之后valid仍为true，则更新在栈中的最大值元素maxval，a出栈并将小于a的所有还没有入栈的元素都压入栈中。另一个解法是考虑每个元素进入栈中只有两种操作：(1)进栈然后出栈(2)进栈然后暂不出栈。也即存在进栈，立即出栈和暂缓出栈三种方式。依次考察出栈序列的每个元素并判断。下面第一份代码是更好的实现
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/************************************************** filename :l.cpp author :maksyuki created time :2018/1/28 9:00:09 last modified :2018/1/28 9:46:33 file location :C:\Users\abcd\Desktop\TheEternalPoet ***************************************************/ #pragma comment(linker, "/STACK:102400000,102400000") #include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> using namespace std; #define CFF freopen ("in", "r", stdin) #define CFO freopen ("out", "w", stdout) #define CPPFF ifstream cin ("in") #define CPPFO ofstream cout ("out") #define DB(ccc) cout << #ccc << " = " << ccc << endl #define DBT printf("time used: %.2lfs\n", (double) clock() / CLOCKS_PER_SEC) #define PB push_back #define MP(A, B) make_pair(A, B) typedef long long LL; typedef unsigned long long ULL; typedef double DB; typedef pair <int, int> PII; typedef pair <int, bool> PIB; const int INF = 0x7F7F7F7F; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 1e5 + 6666; int aa[maxn]; int main() { #ifdef LOCAL CFF; //CFO; #endif int n; while(cin >> n) { if(!n) break; while(cin >> aa[1]) { if(!aa[1]) break; for(int i = 2; i <= n; i++) cin >> aa[i]; stack<int> sta; bool ok = true; int a = 1, p = 1; while(p <= n) { if(a == aa[p]) a++, p++; else if(!sta.empty() && sta.top() == aa[p]) sta.pop(), p++; else if(a <= n) sta.push(a++); else { ok = false; break; } } if(ok) puts("Yes"); else puts("No"); } puts(""); } return 0; } |
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#include <iostream> #include <fstream> #include <cstring> using namespace std; const int maxn = 1000 + 66; int Stack[maxn]; int main() { //ifstream cin ("aaa.txt"); int n; int maxval; while (cin >> n) { if (n == 0) break; int val; while (1) { maxval = 0; bool is_valid = true; memset (Stack, 0, sizeof (Stack)); //0 means no-pushing 1 means push 2 means pop int k; for (k = 0; k < n; k++) { cin >> val; if (val == 0) break; if (is_valid) { int j; for (j= val + 1; j <= maxval; j++) { if (Stack[j] == 1) { is_valid = false; break; } } if (is_valid) { if (maxval < val) maxval = val; Stack[val] = 2; for (j = val - 1; j > 0 && !Stack[j]; j--) Stack[j] = 1; } } } if (is_valid && k == n) cout << "Yes" << endl; else if (!is_valid && k == n) cout << "No" << endl; else { cout << endl; break; } } } return 0; } |
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