uva 112 - Tree Summing

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数据结构。

给一颗树，求根到各个最下面的结点和。有则yes，无则no。

主要是输入的时候不是一行，有空格和回车。

我用栈把树的结点值求存入，然后到最后判断，判断完拿出来，不过貌似处理蛮麻烦的。

#include<cstdio>#include<cstring>#include<stack>using namespace std;int main(){    //freopen("in.txt","r",stdin);    int n;    while(~scanf("%d",&n)){        stack<char>csta;        stack<int>ista;        while(!ista.empty()) ista.pop();        int temp=0,cnt=0,ans=0;        bool ok=false;        bool solve=false;        while(1){            char c=getchar();            if(c<='9' && c>='0'||c=='-'){                temp=0;                bool is=false;                if(c=='-') {                    is = true;                    while(1){                        c = getchar();                        if(c==' '|| c=='\n') continue;                        else break;                    }                }                while(1){                    temp = temp*10 + (c-48);                    c = getchar();                    if(!(c<='9'&&c>='0')) break;                }                if(is) temp = -temp;                ans += temp;                ista.push(temp);                cnt=0;            }            if(c==' ' || c=='\n') continue;            if(c=='('){                ok=false;                csta.push(c);            }            else if(c==')'){                if(!ok) {cnt++;csta.pop();ok=true;}                else{                    temp = ista.top();                    ista.pop();                    ans -= temp;                    csta.pop();                    cnt=0;                }            }            if(cnt==2){                if(ans==n) solve=true;                cnt=0;ok=true;            }            if(csta.empty()) break;        }        if(solve) printf("yes\n");        else printf("no\n");    }    return 0;}

下面某一大神的代码，现在看不懂~，以后看下……和加强版测试数据

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 () () ) ) ) ) )

#include<iostream>#include<cstring>#include<cstdlib>#include<cstdio>#include<cmath>#include<string>using namespace std;int flag;int t_sum(int n,int sum){    int data;    char c;    cin>>c;     //输入左括号    cin>>data;    if(!(cin==0))    {        sum+=data;        int ok1=t_sum(n,sum);  //左树        int ok2=t_sum(n,sum);  //右树        if(!ok1&&!ok2&&!flag)  //如果左右两树都为空，即是叶子，且flag还是0，那么对总和sum与n进行比较            if(sum==n)         //若相等则将flag赋为1                flag=1;        cin>>c;    //输入右括号        return 1;    }    else    {        cin.clear();  //清除错误        cin>>c;        return 0;    }}int main(){    //freopen("sample.txt","r",stdin);    int n;    while(cin>>n)    {        flag=0;        t_sum(n,0);        cout<<(flag?"yes":"no")<<endl;    }    return 0;}

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