poj 1363 火车进站 (栈的应用)
Description
The local tradition is that every train arriving from the direction A continues in the direction B with coaches reorganized in some way. Assume that the train arriving from the direction A has N <= 1000 coaches numbered in increasing order 1, 2, ..., N. The chief for train reorganizations must know whether it is possible to marshal coaches continuing in the direction B so that their order will be a1, a2, ..., aN. Help him and write a program that decides whether it is possible to get the required order of coaches. You can assume that single coaches can be disconnected from the train before they enter the station and that they can move themselves until they are on the track in the direction B. You can also suppose that at any time there can be located as many coaches as necessary in the station. But once a coach has entered the station it cannot return to the track in the direction A and also once it has left the station in the direction B it cannot return back to the station.
Input
The last block consists of just one line containing 0.
Output
Sample Input
5
1 2 3 4 5
5 4 1 2 3
0
6
6 5 4 3 2 1
0
0
Sample Output
Yes
No Yes 先输入火车车厢数N,然后输入N个数代表车厢编号a[i],判断火车是否能按1~N或者N~1的顺序开出,a[0] = 0;代表车厢数为N的循环结束,N = 0;代表程序结束。
思路:
先输入车厢编号,用数组a[i]保存,在从1~N依次输入栈st中,依次判断st.top()是否等于[i].
#include<cstdio>
#include<stack>
using namespace std;
int main()
{
int n,a[],i,j;
while(scanf("%d",&n) && n)
{
stack<int>st;
while(scanf("%d",&a[]) && a[])
{
for( i = ; i < n ; i++)
{
scanf("%d",&a[i]);
}
for( i = , j = ; i <= n ; i++)
{
st.push(i);
while(st.top() == a[j])
{
j++;
st.pop();
if(st.empty()) //栈为空时结束循环
break;
}
}
if( j == n) printf("Yes\n");
else
printf("No\n");
}
printf("\n");
}
}
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