数据结构 uva-11111-Generalized Matrioshkas

题目链接：

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2052



题目意思：

as the problem statement is too long,i haven't read the problem statement Completely during the Contest, consider you have some boxes that some of them are nested in some other boxes , each box should begin with negative
number and end with the absolute value of that negative number , the absolute value shows the size of box , and the sum of size of inside boxes should be less than the outside boxes , you are given sequence of the numbers , you should check
whether it violates the above rules or not, that's all .Hope it helps !



解题思路：

用一个栈模拟，其中遇到负数时进栈，遇到正数处理，如果和当前栈顶元素互补，并且栈顶元素的“和”小于当前正数，则出栈满足要求，并且出栈后将该值加到当前的栈顶的“和”中。详见代码。



代码：

#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#define eps 1e-6
#define INF (1<<20)
#define PI acos(-1.0)
using namespace std;

struct Stack
{
    int top,flag;   //flag记录满不满足题目要求
    int save[200][2];  //save[i][0]表示当前负数的值，save[i][1]表示当前负数所辖领域的总和

    void push(int n)
    {
        ++top;
        save[top][0]=n;  //初始化
        save[top][1]=0;
    }
    int pop()
    {
        return save[top--][0];  //出栈
    }
    bool empty()
    {
        if(top)
            return false;
        return true;
    }
    void set()
    {
        top=flag=0;
    }
}stack;

int main()
{
    int cur;

    while(scanf("%d",&cur)!=EOF)
    {
        stack.set();

        int last=cur;
        char temp=getchar();

        if(cur<0)
            stack.push(cur);
        else     //如果一开始就为正数，肯定不行
            stack.flag=1;

        while(temp!='\n')
        {
            scanf("%d%c",&cur,&temp);

            if(cur<0)   //如果是负数，则直接压栈
                stack.push(cur);
            else
            {
                int temptop=stack.top;

                if(stack.save[temptop][0]+cur==0) //如果与当前栈顶元素互补
                {
                    if(stack.save[temptop][1]<cur) //如果当前栈顶“和”小于当前值，则满足要求
                    {
                        stack.pop();
                        if(!stack.empty())
                            stack.save[stack.top][1]+=cur;
                    }                          //不满足要求
                    else
                        stack.flag=1;
                }
                else   //不满足要求
                    stack.flag=1;
            }
        }
        if(stack.flag==1||!stack.empty())
            printf(":-( Try again.\n");
        else
            printf(":-) Matrioshka!\n");
    }

    return 0;
}