1051. Pop Sequence (25) 时间限制 100 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue
Given a stack which can keep M numbers at most. Push N numbers in the order of 1, 2, 3, …, N and pop randomly. You are supposed to tell if a given sequence of numbers is a possible pop sequence of the stack. For example, if M is 5 and N is 7, we can obtain 1, 2, 3, 4, 5, 6, 7 from the stack, but not 3, 2, 1, 7, 5, 6, 4. Input Specification: Each input file contains one test case. For each case, the first line contains 3 numbers (all no more than 1000): M (the maximum capacity of the stack), N (the length of push sequence), and K (the number of pop sequences to be checked). Then K lines follow, each contains a pop sequence of N numbers. All the numbers in a line are separated by a space. Output Specification: For each pop sequence, print in one line “YES” if it is indeed a possible pop sequence of the stack, or “NO” if not. Sample Input: 5 7 5 1 2 3 4 5 6 7 3 2 1 7 5 6 4 7 6 5 4 3 2 1 5 6 4 3 7 2 1 1 7 6 5 4 3 2 Sample Output: YES NO NO YES NO
#include <iostream> #include <stack> using namespace std; int main() { #ifdef _Debug freopen("data.txt","r+",stdin); #endif // _Debug int M,N,K,cseq[1010]; int cidx; stack <int> se; bool valid; scanf("%d %d %d",&M,&N,&K); while(K--) { valid = true;cidx = 0; while(se.size()) se.pop(); for(int i=0;i<N;++i) scanf("%d",&cseq[i]); for(int i=1;i<=N;) { if(i < cseq[cidx]) { if(se.size() >= M) { valid = false; break; } se.push(i++); } else if(i == cseq[cidx]) { if(se.size() >= M) { valid = false; break; } ++i; ++cidx; } else if(i > cseq[cidx]) { int e = se.top(); se.pop(); if(e != cseq[cidx]) { valid = false; break; } ++cidx; } } while(se.size()) { int e = se.top(); se.pop(); if(e != cseq[cidx++]) { valid = false; break; } } printf("%s\n",valid?"YES":"NO"); } }