POJ1363 Rails【stack】【栈】

标签：des   style   blog   http   io   ar   os   sp   for   

The last block consists of just one line containing 0.

The output contains the lines corresponding to the lines with permutations in the input. A line of the output contains Yes if it is possible to marshal the coaches in the order required on the corresponding line of the input. Otherwise it contains No. In addition, there is one empty line after the lines corresponding to one block of the input. There is no line in the output corresponding to the last ``null‘‘ block of the input.

0

0

Yes

Central Europe 1997

题目大意：如上图所示，已知火车要从A入站，然后从C出站。火车进站的

顺序为1~N，现在给你出站的顺序。问：能不能通过站台改变火车出站顺序

来实现按所给顺序出站。

思路：把站台看做是一个栈，按1~N的顺序遍历火车原先顺序，先入栈，如

果栈顶的火车编号和所给出站顺序将要出站的编号一样。那么火车就出栈，

直到栈里边所有满足出站顺序的火车都出站，否则就一直入栈。最后判断所

有火车是否都出站了。若都出站，输出Yes，否则输出No。

#include<cstdio>
#include<string.h>
#include<stack>
using namespace std;
const int MAXN = 1010;

stack<int> s;
int train[MAXN];

int main()
{
    int N;
    while(~scanf("%d",&N) && N)
    {
        while(scanf("%d",&train[1]) && train[1])
        {
            for(int i = 2; i <= N; i++)
                scanf("%d",&train[i]);
            int B = 1;
            for(int i = 1; i <= N; i++)
            {
                s.push(i);
                while(!s.empty() && s.top() == train[B])
                {
                    s.pop();
                    B++;
                }
            }

            if(B == N+1)
                printf("Yes\n");
            else
                printf("No\n");
            memset(train,0,sizeof(train));
        }
        printf("\n");
    }

    return 0;
}

POJ1363 Rails【stack】【栈】

标签：des   style   blog   http   io   ar   os   sp   for   

原文地址：http://blog.csdn.net/lianai911/article/details/41827347