usaco2.3 zerosum 表达式求值 栈

题目链接

题目大意：

给 1 - n 的数字，在其中添加 {'+', '-', ' '}，使其最终值为0

思路很简单，

因为 n <= 9, 所以可以dfs先生成所有可能的情况组合

然后对于每个组合去判断值是否为 0

因为感觉空格稍稍有点麻烦(...)，就用了栈，还是挺轻松就搞定了的

AC代码如下：

/*
PROB: zerosum
LANG: C++
ID: fan_0111
*/
#include <iostream>
#include <stack>
#include <cstdio>
using namespace std;
char ch[3] = {' ', '+', '-'};
int a[110], n;
stack<int> digit;
stack<char> sign;
bool check() {
	while (!digit.empty()) digit.pop();
	while (!sign.empty()) sign.pop();
	for (int i = 0; i < 2*n; ++i) {
		if (a[i] >= 1 && a[i] <= n) { digit.push(a[i]); continue; }
		if (a[i] == ' ') { 
			if (!sign.empty() && sign.top() == ' ') {
				int x2 = digit.top(); digit.pop();
				int x1 = digit.top(); digit.pop();
				sign.pop();
				digit.push(x1*10 + x2);
			}
			sign.push(a[i]);
			continue;
		}
		while (!sign.empty()) {
			int x2 = digit.top(); digit.pop();
			int x1 = digit.top(); digit.pop();
			int si = sign.top(); sign.pop();
			switch (si) {
				case '+': digit.push(x1 + x2); break;
				case '-': digit.push(x1 - x2); break;
				case ' ': digit.push(x1*10 + x2); break;
			}
		}
		sign.push(a[i]);
	}
	return digit.top() == 0;
}
void print() {
	for (int i = 0; i < 2*n-1; ++i) {
		if (i & 1) cout << (char)a[i];
		else	   cout << a[i];
	}
	cout << endl;
}
void dfs(int dep) {
	if (dep == n-1) {
		if (check()) print();
		return;
	}
	for (int i = 0; i < 3; ++i) {
		a[2*dep+1] = ch[i];
		dfs(dep+1);
	}
}
int main() {
	freopen("zerosum.in", "r", stdin);
	freopen("zerosum.out", "w", stdout);
	cin >> n;
	for (int i = 1; i <= n; ++i) {
		a[i*2-2] = i;
	}
	a[2*n-1] = '+';
	dfs(0);
	fclose(stdin);
	fclose(stdout);
	return 0;
}

但是事实上，后来看 ANALYSIS, 里面的判断是否为 0 的做法还是挺普通的...虽然 switch 用得很好看就是了

不知道我当时为什么没想到...

再看看自己的栈，还是写的挺丑的

ANALYSIS 部分代码如下：

int
eval(char *s)
{
    int term, sign, sum;
    char *p;

    sign = 1;
    term = 0;
    sum = 0;
    for(p=s; *p; p++) {
        switch(*p){
        case '+':
        case '-':
            sum += sign*term;
            term = 0;
            sign = *p == '+' ? 1 : -1;
            break;
        case ' ':
            break;
        default:    /* digit */
            term = term*10 + *p-'0';
        }
    }
    sum += sign*term;
    return sum;
}