栈 紫书ch6例题

两个月前开始看lrj的紫书，看了一个月，大体上过了一遍，但很少自己动手写代码，心想着到学校后重新从头到尾仔细精做一遍，但是还是由于各种事情还是搁了一个月。趁着国庆假期开一个头，学校里这学期教数据结构，于是就决定从第六章开始，一天半除去晚上，不看题解刷了5道题(我好菜啊)。

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贴上刷的两道关于栈的水题，不过还是比较经典的，可以让新手熟悉栈的相关操作：

铁轨

// by cyc
#include<stdio.h>
#include<stack>
#define maxn 1005
using namespace std;

int N, coach[maxn];

void reorganize()
{
	stack<int> s;
	int i = 1, j = 1;
	while (i <= N &&j <= N)
	{
		if (coach[j] == i)
		{ j++; i++; }
		else if (!s.empty() && s.top() == coach[j])
		{ j++; s.pop(); }
		else s.push(i++);
	}
	int ok = 1;
	while (!s.empty())
	{
		if (s.top() != coach[j])
		{
			ok = 0;
			break;
		}
		s.pop(); j++;
	}
	if (ok) printf("Yes\n");
	else printf("No\n");
}

int main()
{
	while (scanf("%d", &N) && N != 0)
	{
		while (scanf("%d", &coach[1]) && coach[1] != 0)
		{
			for (int i = 2; i <= N; i++)
				scanf("%d", &coach[i]);
			reorganize();
		}
		puts("");//there is one empty line after the lines corresponding to one block of the input file
		// 区别于The output of two consecutive cases will be seperated by a blank line
	}
}

// UVa514 Rails
// Rujia Liu
#include<cstdio>
#include<stack>
using namespace std;
const int MAXN = 1000 + 10;

int n, target[MAXN];

int main() {
  while(scanf("%d", &n) == 1) {
    stack<int> s;
    int A = 1, B = 1;
    for(int i = 1; i <= n; i++)
      scanf("%d", &target[i]);
    int ok = 1;
    while(B <= n) {
      if(A == target[B]){ A++; B++; }
      else if(!s.empty() && s.top() == target[B]){ s.pop(); B++; }
      else if(A <= n) s.push(A++);
      else { ok = 0; break; }
    }
    printf("%s\n", ok ? "Yes" : "No");
  }
  return 0;
}

矩阵链乘

// by cyc
#include<stdio.h>
#include<stack>
#include<vector>
using namespace std;
const int maxn = 30;
typedef pair<int, int> pp;
vector<pp> matrix;

inline int ID(char c)
{
	return c - 'A';
}

int main()
{
	int n;
	char c, expr[30*100];
	for (int i = 0; i < 26; i++)
		matrix.push_back(make_pair(0, 0));
	scanf("%d", &n);
	for (int i = 0; i < n; i++)
	{
		getchar();
		scanf("%c", &c);
		scanf("%d%d", &matrix[ID(c)].first, &matrix[ID(c)].second);
	}
	while (scanf("%s", expr) != EOF)
	{
		stack<char> s;
		int i = 0, answer = 0;
		while (expr[i] != '\0')
		{
			if (expr[i] != ')') s.push(expr[i]);
			else
			{
				int temp = s.top(); s.pop();
				int l = matrix[ID(temp)].first, r = matrix[ID(temp)].second;
				while (s.top() != '(')
				{
					int p = ID(s.top()); s.pop();
					if (matrix[p].second == l)
					{
						answer += matrix[p].first * l * r;
						l = matrix[p].first;
					}
					else
					{
						answer = -1;
						break;
					}
				}
				if (answer == -1)
				{
					printf("error\n");
					break;
				}
				s.pop();
				matrix.push_back(make_pair(l, r));
				s.push(matrix.size()-1+'A');
			}
			i++;
		}
		if (answer!=-1) printf("%d\n", answer);
	}
}

// UVa442 Matrix Chain Multiplication
// Rujia Liu
// 题意：输入n个矩阵的维度和一些矩阵链乘表达式，输出乘法的次数。假定A和m*n的，B是n*p的，那么AB是m*p的，乘法次数为m*n*p
// 算法：用一个栈。遇到字母时入栈，右括号时出栈并计算，然后结果入栈。因为输入保证合法，括号无序入栈
#include<cstdio>
#include<stack>
#include<iostream>
#include<string>
using namespace std;

struct Matrix {
  int a, b;
  Matrix(int a=0, int b=0):a(a),b(b) {}
} m[26];

stack<Matrix> s;

int main() {
  int n;
  cin >> n;
  for(int i = 0; i < n; i++) {
    string name;
    cin >> name;
    int k = name[0] - 'A';
    cin >> m[k].a >> m[k].b;
  }
  string expr;
  while(cin >> expr) {
    int len = expr.length();
    bool error = false;
    int ans = 0;
    for(int i = 0; i < len; i++) {
      if(isalpha(expr[i])) s.push(m[expr[i] - 'A']);
      else if(expr[i] == ')') {
        Matrix m2 = s.top(); s.pop();
        Matrix m1 = s.top(); s.pop();
        if(m1.b != m2.a) { error = true; break; }
        ans += m1.a * m1.b * m2.b;
        s.push(Matrix(m1.a, m2.b));        
      }
    }
    if(error) printf("error\n"); else printf("%d\n", ans);
  }

  return 0;
}

虽说是两道水题，但是代码实现还是不够凝练，和lrj的差距简直不要太远，还需多总结提高