poj3195 Generalized Matrioshkas(栈)

Description

Vladimir worked for years making matrioshkas, those nesting dolls that certainly represent truly Russian craft. A matrioshka is a doll that may be opened in two halves, so that one finds another doll inside. Then this doll may be opened to find another one inside it. This can be repeated several times, till a final doll – that cannot be opened – is reached.

Recently, Vladimir realized that the idea of nesting dolls might be generalized to nesting toys. Indeed, he has designed toys that contain toys but in a more general sense. One of these toys may be opened in two halves and it may have more than one toy inside it. That is the new feature that Vladimir wants to introduce in his new line of toys.

Vladimir has developed a notation to describe how nesting toys should be constructed. A toy is represented with a positive integer, according to its size. More precisely: if when opening the toy represented by m we find the toys represented by n1, n2, …, nr, it must be true that n1 + n2 + … + nr < m. And if this is the case, we say that toy m contains directly the toys n1, n2, …, nr. It should be clear that toys that may be contained in any of the toys n1, n2, …, nr are not considered as directly contained in the toy m.

A generalized matrioshka is denoted with a non-empty sequence of non zero integers of the form:

a1 a2 … aN
such that toy k is represented in the sequence with two integers −k and k, with the negative one occurring in the sequence first that the positive one.

For example, the sequence

−9 −7 −2 2 −3 −2 −1 1 2 3 7 9
represents a generalized matrioshka conformed by six toys, namely, 1, 2 (twice), 3, 7 and 9. Note that toy 7 contains directly toys 2 and 3. Note that the first copy of toy 2 occurs left from the second one and that the second copy contains directly a toy 1. It would be wrong to understand that the first −2 and the last 2 should be paired.

On the other hand, the following sequences do not describe generalized matrioshkas:

−9 −7 −2 2 −3 −1 −2 2 1 3 7 9
because toy 2 is bigger than toy 1 and cannot be allocated inside it.

−9 −7 −2 2 −3 −2 −1 1 2 3 7 −2 2 9
because 7 and 2 may not be allocated together inside 9.

−9 −7 −2 2 −3 −1 −2 3 2 1 7 9
because there is a nesting problem within toy 3.

Your problem is to write a program to help Vladimir telling good designs from bad ones.

Input

The input file contains several test cases, each one of them in a separate line. Each test case is a sequence of non zero integers, each one with an absolute value less than 107.

Output

Output texts for each input case are presented in the same order that input is read.

For each test case the answer must be a line of the form

:-) Matrioshka!

if the design describes a generalized matrioshka. In other case, the answer should be of the form

:-( Try again.

Sample Input
-9 -7 -2 2 -3 -2 -1 1 2 3 7 9
-9 -7 -2 2 -3 -1 -2 2 1 3 7 9
-9 -7 -2 2 -3 -1 -2 3 2 1 7 9
-100 -50 -6 6 50 100
-100 -50 -6 6 45 100
-10 -5 -2 2 5 -4 -3 3 4 10
-9 -5 -2 2 5 -4 -3 3 4 9

Sample Output
:-) Matrioshka!
:-( Try again.
:-( Try again.
:-) Matrioshka!
:-( Try again.
:-) Matrioshka!
:-( Try again.

分析：
题意就是一些俄罗斯套娃
满足没有大的装小的，没有互相交叉，
每一个娃娃（设为A）里装的所有小娃娃的大小加起来小于A的大小
判断是否合法

这道题一看上去超蒙的
一开始的想法，可以O（n）扫一遍，一旦遇到这样的：-a a，
我们就进行合并并且记录套娃的大小
当没法合并的时候，我们继续进行遍历

在手玩的时候，发现这种数据结构有点像栈，
我实际上维护了两个栈

tip

题目没有明确的数据范围，坑。。。
提交的时候要用G++

这里写代码片
#include<cstdio>
#include<cstring>
#include<iostream>

using namespace std;

const int N=1000001;
int n,m;
char s;
int a[N],sta[N][2],top,top2,st[N][2];

int abs(int x) {
    if (x>0) return x;
    else return -x;
}

int doit()
{
    int i,j;
    top=0; top2=0;
    for (int i=1;i<=n;i++)
    {
        int cnt=0;
        if (top&&sta[top][0]==-a[i]){   //要合并 
            while (top2&&st[top2][1]>sta[top][1])   //包含 
            {
                cnt+=st[top2][0];
                top2--;
            }
            if (abs(a[i])>cnt) st[++top2][0]=abs(a[i]),st[top2][1]=sta[top][1];
            else return 0;
            top--;
        } 
        else{
            sta[++top][0]=a[i]; sta[top][1]=i;   //1  编号 
        }
    }
    if (top2==1&&top==0) return 1;
    else return 0; 
}

int main()
{
    while (scanf("%c",&s)!=EOF)
    {
        n=0;
        int ff=1;
        int num=0;
        while (s!='
')
        {
            if ((s<'0'||s>'9')&&s=='-') ff=-1;
            if (s>='0'&&s<='9') num=num*10+s-'0';
            if (s==' ') a[++n]=num*ff,num=0,ff=1; 
            scanf("%c",&s);
        }
        a[++n]=ff*num;
        if (doit()) printf(":-) Matrioshka!
");
        else printf(":-( Try again.
");
    }
    return 0;
}