5-3 Pop Sequence (25分) / PAT 1051. Pop Sequence (25)

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5-3 Pop Sequence (25分)

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 51 2 3 4 5 6 73 2 1 7 5 6 47 6 5 4 3 2 15 6 4 3 7 2 11 7 6 5 4 3 2

Sample Output:

YESNONOYESNO

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

http://pta.patest.cn/pta/test/16/exam/4/question/665

#include <cstdio>  #include <sstream>  #include <cstring>  #include <iostream>#include <string>#include <vector>#include <algorithm>#include <stack>using namespace std;#define  N 1001/*M (the maximum capacity of the stack)N (the length of push sequence)*/int m, n, kk;int a[N];int main(){    //freopen("in.txt", "r", stdin);    scanf("%d%d%d", &m, &n, &kk);    int i, j;    while (kk--)    {        for (j = 0; j < n; j++)        {            scanf("%d", &a[j]);        }if(a[0] > m) // 这里第一个元素就比栈容量还大 直接 NO 跳过{printf("NO\n");continue;}stack<int> sta ;int nextNum ;for(i = 1 ; i <= a[0]  ; i++){sta.push(i) ;}int popNum = sta.top() ;sta.pop(); // 弹出第一个数nextNum = popNum + 1 ; // 下一个应该入栈的数bool flag = true ;for(i = 1 ; i < n ; i++){int nowNum = a[i] ;if(nowNum < nextNum){if(sta.empty()){flag = false ;}else{int staTop = sta.top() ;if(nowNum < staTop) // 小于栈顶元素 错误{flag = false ;break ;}else if(nowNum == staTop){ // 等于栈顶元素 直接弹出sta.pop() ;}}if(flag == false)break;}else{ // 大于等于下一个该入栈的元素int cnt = nowNum - nextNum + 1 ;int cnt2 = cnt + (int)sta.size() ;if(cnt2 > m) // 栈溢出 {flag = false ;break;}for(j = nextNum ; j <= nowNum ; j++){sta.push(j) ;}sta.pop() ; // 弹出该元素nextNum = nowNum + 1; //下一个该入栈的元素 为此元素加上1 }}if(flag == false){printf("NO\n") ;}else{printf("YES\n") ;}}return 0 ;}

第二次ac,注意变量含义，防止溢出等不合理现象产生

#include <cstdio>#include <cstdlib>#include <cmath>#include <cstring>#include <iostream>#include <string>#include <vector>#include <queue>#include <algorithm>#include <sstream>#include <stack> #include <map> #include <set> #include <unordered_map>using namespace std;const int INF = 0x7fffffff;const int MIN_INF = - INF -1;typedef long long int LL;int m,n,k;int main(){//freopen("in.txt","r",stdin);scanf("%d%d%d", &m,&n,&k);while(k--){vector<int> v(n,0);for(int j=0; j<n; j++){scanf("%d", &v[j]);}if(v[0] > m){printf("NO\n");continue;}stack<int> sta;for(int i=1;i<=v[0];i++){sta.push(i);}sta.pop();int nextP = v[0] + 1;bool flag = true;for(int i=1;i<n;i++){if(v[i] == nextP){int len = sta.size();if(len >= m){flag = false;break;}nextP = nextP + 1;}else if(v[i] < nextP){if(sta.empty()){flag = false;break;}int topVal = sta.top();while(topVal != v[i]){sta.pop();if(sta.empty())break;topVal = sta.top();}if(!sta.empty() && sta.top() == v[i])sta.pop();else{flag = false;break;}}else{for(int j=nextP ;j<=v[i];j++){sta.push(j);}nextP = v[i] + 1;int len = sta.size();if(len > m){flag = false;break;}else{if(!sta.empty())sta.pop();}}}if(flag){printf("YES\n");}else{printf("NO\n");}}//printf("\n");return 0;}

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