HDU 1022 Train Problem I (STL 棧類比)

標籤：

Train Problem ITime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 30420    Accepted Submission(s): 11492


Problem DescriptionAs the new term comes, the Ignatius Train Station is very busy nowadays. A lot of student want to get back to school by train(because the trains in the Ignatius Train Station is the fastest all over the world ^v^). But here comes a problem, there is only one railway where all the trains stop. So all the trains come in from one side and get out from the other side. For this problem, if train A gets into the railway first, and then train B gets into the railway before train A leaves, train A can‘t leave until train B leaves. The pictures below figure out the problem. Now the problem for you is, there are at most 9 trains in the station, all the trains has an ID(numbered from 1 to n), the trains get into the railway in an order O1, your task is to determine whether the trains can get out in an order O2.
 
InputThe input contains several test cases. Each test case consists of an integer, the number of trains, and two strings, the order of the trains come in:O1, and the order of the trains leave:O2. The input is terminated by the end of file. More details in the Sample Input.
 
OutputThe output contains a string "No." if you can‘t exchange O2 to O1, or you should output a line contains "Yes.", and then output your way in exchanging the order(you should output "in" for a train getting into the railway, and "out" for a train getting out of the railway). Print a line contains "FINISH" after each test case. More details in the Sample Output.
 
Sample Input

3 123 3213 123 312

 
Sample Output

Yes.inininoutoutoutFINISHNo.FINISHHintHint For the first Sample Input, we let train 1 get in, then train 2 and train 3.So now train 3 is at the top of the railway, so train 3 can leave first, then train 2 and train 1.In the second Sample input, we should let train 3 leave first, so we have to let train 1 get in, then train 2 and train 3.Now we can let train 3 leave.But after that we can‘t let train 1 leave before train 2, because train 2 is at the top of the railway at the moment.So we output "No.".

 
Author

Ignatius.L

原題連結:http://acm.hdu.edu.cn/showproblem.php?pid=1022

題意:或者站只有一條鐵軌,問火車是否可以按前面的序列進站後是否可以按後面的序列出站.

棧類比:

AC代碼:

#include<iostream>#include<stack>#define maxn 100using namespace std;int main(){    stack<char>s;    int n,i,j,k;    bool result[maxn];//result數組用來表示結果，1表示進棧。0表示出    char in[maxn],out[maxn];//序列1和序列2    while(cin>>n>>in>>out)    {        j=0,i=0,k=1;        s.push(in[0]);//為防止棧空，壓一個進去        result[0]=1;//記錄進來了一個。        while(i<n&&j<n)        {            if(!s.empty()&&s.top()==out[j])            {                //如果棧頂元素與序列2當前的元素相等，則彈棧，序列2集團向後移一位。                j++;                s.pop();                result[k++]=0;            }            else            {                //否則從序列1中取當前元素壓入棧中。                s.push(in[++i]);                result[k++]=1;            }        }        if(i==n)//如果I==N表示棧頂元素不等於序列2當前元素，且序列1中元素都已經入過棧，判斷不能得到序列2一樣的答案。            cout<<"No."<<endl;        else        {            cout<<"Yes."<<endl;            for(i=0; i<k; i++)                if(result[i])                    cout<<"in"<<endl;                else                    cout<<"out"<<endl;        }        cout<<"FINISH"<<endl;    }    return 0;}

AC代碼2:

#include <iostream>#include <stack>#include <cstdio>#include <cstring>using namespace std;stack<int> S;int main(){    int n,i,j,k;    bool flag[10000];    string in,out;    while(~scanf("%d%*c",&n))    {        cin >> in >> out;        i = j = k = 0;        while(j<n+1 && i<n)        {            if(!S.empty() && S.top() == out[i])            {                i++;                flag[k] = true;                k++;                S.pop();            }            else            {                S.push(in[j]);                flag[k] = false;                j++;                k++;            }        }        cout<<k<<endl;        if(k!=2*n)        {            cout << "No." << endl;        }        else        {            cout << "Yes." << endl;            for(i = 0; i<k; i++)            {                if(flag[i])                    cout << "out" << endl;                else                    cout << "in" << endl;            }        }        cout << "FINISH" << endl;    }    return 0;}

AC代碼3:C語言類比:

# include<stdio.h># include<string.h>int main(){    int n,i,j,k,head;    char a[1000],b[1000],t[1000],str[1000][10];    while(scanf("%d%s%s",&n,a,b)!=EOF)    {        head=i=j=k=0;        strcpy(str[k++],"in");        while(i<n)        {            t[head]=a[i];            while(t[head]==b[j]&&b[j]!='\0'&&t[head]!='\0')            {                head--;                j++;                strcpy(str[k++],"out");            }            head++;            strcpy(str[k++],"in");            i++;        }        if(head==0)        {            printf("Yes.\n");            for(i=0; i<k-1; i++)                printf("%s\n",str[i]);            printf("FINISH\n");        }        else        {            printf("No.\n");            printf("FINISH\n");        }    }    return 0;}

HDU 1022 Train Problem I (STL 棧類比)