HDU-3350 #define is unsafe（dfs 解析字符串）

本文链接：https://blog.youkuaiyun.com/zzcblogs/article/details/73136656

本文通过对比宏定义与函数在实现最大值选择上的差异，揭示了宏定义可能导致重复计算的问题，并提供了一种通过深度优先搜索统计计算次数的方法。

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#define is unsafe

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 75 Accepted Submission(s): 57

Problem Description

Have you used #define in C/C++ code like the code below?

#include <stdio.h>
#define MAX(a , b) ((a) > (b) ? (a) : (b))
int main()
{
  printf("%d\n" , MAX(2 + 3 , 4));
  return 0;
}

Run the code and get an output: 5, right?
You may think it is equal to this code:

#include <stdio.h>
int max(a , b) {  return ((a) > (b) ? (a) : (b));  }
int main()
{
  printf("%d\n" , max(2 + 3 , 4));
  return 0;
}

But they aren't.Though they do produce the same anwser , they work in two different ways.
The first code, just replace the MAX(2 + 3 , 4) with ((2 + 3) > (4) ? (2 + 3) : 4), which calculates (2 + 3) twice.
While the second calculates (2 + 3) first, and send the value (5 , 4) to function max(a , b) , which calculates (2 + 3) only once.

What about MAX( MAX(1+2,2) , 3 ) ?
Remember "replace".
First replace: MAX( (1 + 2) > 2 ? (1 + 2) : 2 , 3)
Second replace: ( ( ( 1 + 2 ) > 2 ? ( 1 + 2 ) : 2 ) > 3 ? ( ( 1 + 2 ) > 2 ? ( 1 + 2 ) : 2 ) : 3).
The code may calculate the same expression many times like ( 1 + 2 ) above.
So #define isn't good.In this problem,I'll give you some strings, tell me the result and how many additions(加法) are computed.

Input

The first line is an integer T(T<=40) indicating case number.
The next T lines each has a string(no longer than 1000), with MAX(a,b), digits, '+' only(Yes, there're no other characters).
In MAX(a,b), a and b may be a string with MAX(c,d), digits, '+'.See the sample and things will be clearer.

Output

For each case, output two integers in a line separated by a single space.Integers in output won't exceed 1000000.

Sample Input

6
MAX(1,0)
1+MAX(1,0)
MAX(2+1,3)
MAX(4,2+2)
MAX(1+1,2)+MAX(2,3)
MAX(MAX(1+2,3),MAX(4+5+6,MAX(7+8,9)))+MAX(10,MAX(MAX(11,12),13))

Sample Output

Author

madfrog

Source

HDU2010省赛集训队选拔赛（校内赛）

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lcy

把一个括号中的 lvalue, rvalue分成两个实例进行dfs 非常巧妙

#include <iostream>
#include <algorithm>
#include <array>
#include <cmath>
#include <stdio.h>
using namespace std;
struct node {
    int value, cnt;//值 跟 被引用的次数
    node(int va = 0, int cn = 0) : value(va), cnt(cn){}
};
string str;
int dfs(int index, node& n) {
    int i, num = 0;
    for (i = index; i < str.size();) {
        if (str[i] == '(') {//解析一个括号
            node left, right;
            i = dfs(i + 1, left);// 遇到"(" 加一
            i = dfs(i + 1, right);// 遇到"," 加一
            if (left.value > right.value) {
                n.cnt += 2*left.cnt + right.cnt;
                n.value += left.value;
            } else {
                n.cnt += 2*right.cnt + left.cnt;
                n.value += right.value;
            }
            //遇到 "(" 加一
            num = 0;
            i++;
        } else if (str[i] == ')' || str[i] == ',') {
            n.value += num;
            return i;
        } else if (str[i] == '+') {
            n.value += num;
            num = 0;
            n.cnt++;
            ++i;
        } else if (str[i] >= '0' && str[i] <= '9') {
            num = num * 10 + (str[i] - '0');
            ++i;
        } else
            ++i;
    }
    if (num)//加法在括号外的值也要加上
        n.value += num;
    return i;
}
int main () {
    int n;
    cin >> n;
    while (n--) {
        node p;
        cin >> str;
        dfs(0, p);
        cout << p.value << " " << p.cnt << endl;
    }
}