POJ 1068 解题报告

来源：互联网发布：apache官网下载教程编辑：程序博客网时间：2024/06/16 22:27

这道题就是按照题意模拟做出来的，主要是数据量很小。通过输入的p可以确定括号序列，然后按照次序注意处理，左括号压栈，右括号入栈，同时统计w值。

1068Accepted248K16MSC++1314B

/* ID: thestor1 LANG: C++ TASK: poj1068 */#include <iostream>#include <fstream>#include <cmath>#include <cstdio>#include <cstring>#include <limits>#include <string>#include <vector>#include <list>#include <set>#include <map>#include <queue>#include <stack>#include <algorithm>#include <cassert>using namespace std;const int MAXN = 100;int main(){std::ios::sync_with_stdio(false);std::vector<int> parentheses(MAXN, 0);int T;cin >> T;for (int t = 0; t < T; ++t){int n;cin >> n;int k = 0, p0 = 0;for (int i = 0; i < n; ++i){int p;cin >> p;for (int j = 0; j < p - p0; ++j){parentheses[k] = 1;k++;}parentheses[k] = 0;k++;p0 = p;}n = 2 * n;// cout << "parentheses:";// for (int i = 0; i < n; ++i)// {// cout << parentheses[i];// }// cout << endl;stack<int> st;for (int i = 0; i < n; ++i){// left parentheseif (parentheses[i]){st.push(i);}else{int left = st.top();st.pop();int cnt = 1;for (int j = left + 1; j < i; ++j){if (parentheses[j] == 0){cnt++;}}cout << cnt << " ";}}cout << endl;}return 0;  }

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