UVa 514 - Rails

https://uva.onlinejudge.org/external/5/514.pdf

514 Rails

There is a famous railway station in PopPush City. Country there is incredibly hilly. The station was

built in last century. Unfortunately, funds were extremely limited that time. It was possible to establish

only a surface track. Moreover, it turned out that the station could be only a dead-end one (see picture)

and due to lack of available space it could have only one track.

A

5, 4, 3, 2, 1 1, 2, 3, 4, 5

B

Station

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 input le consists of blocks of lines. Each block except the last describes one train and possibly

more requirements for its reorganization. In the rst line of the block there is the integer N described

above. In each of the next lines of the block there is a permutation of 1; 2; : : : ; N The last line of the

block contains just 0.

The last block consists of just one line containing 0.

Output

The output le contains the lines corresponding to the lines with permutations in the input le. A

line of the output le contains Yes if it is possible to marshal the coaches in the order required on the

corresponding line of the input le. Otherwise it contains No. In addition, there is one empty line after

the lines corresponding to one block of the input le. There is no line in the output le corresponding

to the last \null” block of the input le.

ACM Contest Problems Archive University of Valladolid (SPAIN)

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

分析：

简单的把栈给模拟一下就可以了

//STL写法
#include <iostream>
#include <sstream>
#include <iomanip>
#include <vector>
#include <deque>
#include <list>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <bitset>
#include <string>
#include <numeric>
#include <algorithm>
#include <functional>
#include <iterator>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <complex>
#include <ctime>
#define INF 0x3f3f3f3f
#define eps 1e-6

typedef long long LL;
const double pi = acos(-1.0);
const long long mod = 1e9 + 7;
using namespace std;

int target[1005];
stack <int> a;

int main()
{
//freopen("int.txt","r",stdin);
//freopen("out.txt","w",stdout);
int n;
while(scanf("%d",&n) && n)
{
while(scanf("%d",&target[0]) && target[0] != 0)
{
for(int i = 1;i < n;i++)
scanf("%d",&target[i]);
int ok = 1;
int A = 1,B = 0;
while(B < n)
{
if(A == target[B])
{
A++;
B++;
}
else if(!a.empty() && a.top() == target[B])
{
a.pop();
B++;
}
else if(A <= n)
a.push(A++);
else
{
ok = 0;
break;
}
}
if(ok)
puts("Yes");
else
puts("No");
}
puts("");
}
return 0;
}

//数组写法
#include <iostream>
#include <sstream>
#include <iomanip>
#include <vector>
#include <deque>
#include <list>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <bitset>
#include <string>
#include <numeric>
#include <algorithm>
#include <functional>
#include <iterator>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <complex>
#include <ctime>

typedef long long LL;
const double pi = acos(-1.0);
const long long mod = 1e9 + 7;
using namespace std;

int target[1005];
int a[1005];

int main()
{
//freopen("int.txt","r",stdin);
//freopen("out.txt","w",stdout);
int n;
while(scanf("%d",&n) && n)
{
while(scanf("%d",&target[0]) && target[0] != 0)
{
for(int i = 1;i < n;i++)
scanf("%d",&target[i]);
int top = -1,ok = 1;
int A = 1,B = 0;
while(B < n)
{
if(A == target[B])
{
A++;
B++;
}
else if(top >= 0 && a[top] == target[B])
{
top--;
B++;
}
else if(A <= n)
{
a[++top] = A++;
}
else
{
ok = 0;
break;
}
}

if(ok)
puts("Yes");
else
puts("No");
}
puts("");
}
return 0;
}