02-线性结构4 Pop Sequence (25分)

02-线性结构4 Pop Sequence   (25分)

Given a stack which can keep MM numbers at most. Push NN numbers in the order of 1, 2, 3, ..., NN 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 MM is 5 and NN 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): MM (the maximum capacity of the stack), NN (the length of push sequence), and KK (the number of pop sequences to be checked). Then KK lines follow, each contains a pop sequence of NN 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
 
时间限制：400ms
内存限制：64MB
代码长度限制：16kB
判题程序：系统默认
作者：陈越
单位：浙江大学
 

题目判定

题意：栈的大小M，输入序列的长度N（默认序列即为1,2,3…,N），入栈出栈的顺序不定。有K个测试序列，判断每一个测试序列是否是可能的出栈顺序。

分析：就是从出栈顺序退出进栈顺序。当我们遇见输出x时，则要考虑的是x前的元素，即小于等于x的元素都先push栈，才会有pop x；1。栈为空时，判断需要填入的数 是否小于 栈的容量(即M)2。若后一个数比前一个数大，又要push其之前的数 再判断　3。若后一个数比前一个数小，则要判断栈顶元素是否与其相等

#include <stdio.h>#include <iostream>#include <algorithm>#include <vector>#include <stack>using namespace std;int M, N, K;int Check(vector<int> &v){    int i=0;    int num=1;    int cap=M+1;    stack<int> sta;    sta.push(0);    while(i<N)    {        while(v[i]>sta.top() && sta.size()<cap)            sta.push(num++);        if(v[i++]==sta.top())            sta.pop();        else            return 0;    }    return 1;}int main(){    vector<int> vec(N,0);    scanf("%d%d%d",&M,&N,&K);    for(int i=0; i<K; i++)    {        for(int j=0; j<N; j++)        {            int number;            scanf("%d",&number);            vec.push_back(number);        }        if(Check(vec))            printf("YES\n");        else            printf("NO\n");        vec.clear();    }    return 0;}