UVA - 112 - Tree Summing （数的求和！栈的应用！）

UVA - 112

Tree Summing

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

Submit Status

Description

Tree Summing

Background

LISP was one of the earliest high-level programming languages and, with FORTRAN, is one of the oldest languages currently being used. Lists, which are the fundamental data structures in LISP, can easily be adapted
to represent other important data structures such as trees.
This problem deals with determining whether binary trees represented as LISP S-expressions possess a certain property.

The Problem

Given a binary tree of integers, you are to write a program that determines whether there exists a root-to-leaf path whose nodes sum to a specified integer. For example, in the tree shown below there are exactly
four root-to-leaf paths. The sums of the paths are 27, 22, 26, and 18.



Binary trees are represented in the input file as LISP S-expressions having the following form.

empty tree 		 ::= 		 ()
tree ::= empty tree

 (integer treetree)

The tree diagrammed above is represented by the expression (5 (4 (11 (7 () ()) (2 () ()) ) ()) (8 (13 () ()) (4 () (1 () ()) ) ) )
Note that with this formulation all leaves of a tree are of the form (integer () () )
Since an empty tree has no root-to-leaf paths, any query as to whether a path exists whose sum is a specified integer in an empty tree must be answered negatively.

The Input

The input consists of a sequence of test cases in the form of integer/tree pairs. Each test case consists of an integer followed by one or more spaces followed by a binary tree formatted as an S-expression as
described above. All binary tree S-expressions will be valid, but expressions may be spread over several lines and may contain spaces. There will be one or more test cases in an input file, and input is terminated by end-of-file.

The Output

There should be one line of output for each test case (integer/tree pair) in the input file. For each pair I,T (I represents the integer, T represents the tree) the output
is the string yes if there is a root-to-leaf path in T whose sum is I and no if there is no path in T whose sum is I.

Sample Input

22 (5(4(11(7()())(2()()))()) (8(13()())(4()(1()()))))
20 (5(4(11(7()())(2()()))()) (8(13()())(4()(1()()))))
10 (3 
     (2 (4 () () )
        (8 () () ) )
     (1 (6 () () )
        (4 () () ) ) )
5 ()

Sample Output

yes
no
yes
no

Source

Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: Graph :: Special Graphs (Others) :: Tree

Root :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: Graph :: Special Graphs (Others) :: Tree

Root :: Competitive Programming: Increasing the Lower Bound of Programming Contests (Steven & Felix Halim) :: Chapter 4. Graph :: Special
Graph - Tree

Root :: AOAPC I: Beginning Algorithm Contests (Rujia Liu) :: Volume 2. Data Structures :: Binary Trees

我尼玛！这题真做的我整个人都不好了........

首先数据很坑！！

输入：

22 (5(4(11(7()())(2()()))()) (8(13()())(4()(1()()))))

20 (5(4(11(7()())(2()()))()) (8(13()())(4()(1()()))))

10 (3

(2 (4 () () )

(8 () () ) )

(1 (6 () () )

(4 () () ) ) )

5 ()

0 ()

5 (5 () ())

5 ( 5 () () )

5 (1 (3 () ()) (4 () ()))

5 (18 ( - 13 ( ) ( ))())

0 (1 ()(-2 () (1()()) ) )

2 (1 () (1 () (1 () () ) ) )

10 (5 () (5 () (5 () (5 () (4 () () ) ) ) ) )

10 (5 () (5 () (5 () (5 ( 3 () () ) (4 () () ) ) ) ) )

20 (5 () (5 () (5 () (5 () (4 () () ) ) ) ) )



输出：

yes

no

yes

no

no

yes

yes

yes

yes

yes

no

no

no

no



居然尼玛有负号！！

还居然负号和数字之间有空格！！

我差点笑出声啊！！

哈哈哈哈哈%>_<%！！

AC代码：

#include <cstdio> 
#include <cstring>
#include <algorithm>
#include <string>
#include <stack>
#include <queue>
#include <cmath>
using namespace std;

int main()
{
	int n;
	char a[300];
	while(scanf("%d", &n) != EOF)
	{
		stack<int> snum;
		stack<char> sch;
		int sum = 0, flag = 0;
		int iswei = 0;  //判断是否到了叶子节点 
		gets(a);
		while(1)
		{
			for(int i=0; a[i] != '\0'; i++)
			{
				if( (a[i] <= '9' && a[i] >= '0') || a[i] == '-')
				{
					iswei = 0;
					int t = 0, f = 0;
					if(a[i] == '-')
					{
						while(a[i]>'9' || a[i] < '0') i++;
						f = 1;
					}
					while(a[i] <= '9' && a[i] >='0')
					{
						t = t*10 + a[i]-'0';
						i++;
					}
					i--;
					if(f) t = -t;
					sum += t;
					snum.push(t);
				}
				else if(a[i] == '(')
				{
					iswei++;
					sch.push(a[i]);
				}
				else if(a[i] == ')')
				{
					//printf("%d %d\n", sch.size(), snum.size());
					if(sch.size() == snum.size())
					{
						if(iswei == 2 && sum == n)  flag = 1;  //之前在这里就直接break了，导致后面没输入进去，纠结我好久 
						sum -= snum.top();
						snum.pop();
						iswei = 0;
					}
					sch.pop();
					//printf("%d\n", sum);
				}
			}
			if(sch.empty()) break;   //当所有（）都匹配好了后即栈为空就说明树输出完毕，然后就跳出循环 
			else 
			{
				gets(a);
			}
		}
		if(flag == 1) printf("yes\n");
		else printf("no\n");
	}
	return 0;
}

另附大牛的代码（Orz...）：

#include<iostream>
using namespace std;
bool ok;
bool tree_sum(int n,int sum)
{
    int v;
    char ch;
    cin>>ch;
    if(!((cin>>v)==0))
    {
        n+=v;
        bool t=tree_sum(n,sum)|tree_sum(n,sum);
        if(!ok&&!t)
            ok=(n==sum);
        cin>>ch;
        return true;
    }
    else
    {
        cin.clear();
        cin>>ch;
        return false;
    }
}
int main()
{
    // freopen("f:\\out.txt", "w", stdout);
    int sum;
    while(cin>>sum)
    {
        ok=false;
        tree_sum(0,sum);
        cout<<(ok?"yes":"no")<<endl;
    }
    return 0;
}