[2016-02-03][UVA][514][Rails]
- 时间:2016-02-03 22:24:52 星期三
- 题目编号:UVA 514
- 题目大意:给定若干的火车(编号1-n),按1-n的顺序进入车站,
- 给出火车出站的顺序,问是否有可能存在
- 分析: FIFO,用栈模拟一遍即可,
- 方法: 根据输入的顺序,从1-n开始,当前操作的为i
- 如果i是当前对应的编号,那么直接跳过(进入B)
- 如果不是,根据当前需求的编号,小于i,就从栈顶弹出一个元素,
- 看这个元素是否是需求的,是则继续.否则NO
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#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
using
namespace
std;
typedef
long
long
LL;
#define CLR(x,y) memset((x),(y),sizeof((x)))
#define getint(x) int (x);scanf("%d",&(x))
#define get2int(x,y) int (x),(y);scanf("%d%d",&(x),&(y))
#define get3int(x,y,z) int (x),(y),(z);scanf("%d%d%d",&(x),&(y),&(z))
#define getll(x) LL (x);scanf("%I64d",&(x))
#define get2ll(x,y) LL (x),(y);scanf("%I64d%I64d",&(x),&(y))
#define get3ll(x,y,z) LL (x),(y),(z);scanf("%I64d%I64d%I64d",&(x),&(y),&(z))
#define getdb(x) double (x);scanf("%lf",&(x))
#define get2db(x,y) double (x),(y);scanf("%lf%lf",&(x),&(y))
#define get3db(x,y,z) double (x),(y),(z);scanf("%lf%lf%lf",&(x),&(y),&(z))
#define getint2(x) scanf("%d",&(x))
#define get2int2(x,y) scanf("%d%d",&(x),&(y))
#define get3int2(x,y,z) scanf("%d%d%d",&(x),&(y),&(z))
#define getll2(x) scanf("%I64d",&(x))
#define get2ll2(x,y) scanf("%I64d%I64d",&(x),&(y))
#define get3ll2(x,y,z) scanf("%I64d%I64d%I64d",&(x),&(y),&(z))
#define getdb2(x) scanf("%lf",&(x))
#define get2db2(x,y) scanf("%lf%lf",&(x),&(y))
#define get3db2(x,y,z) scanf("%lf%lf%lf",&(x),&(y),&(z))
#define getstr(str) scanf("%s",str)
#define get2str(str1,str2) scanf("%s%s",str1,str2)
#define FOR(x,y,z) for(int (x)=(y);(x)<(z);(x)++)
#define FORD(x,y,z) for(int (x)=(y);(x)>=(z);(x)--)
#define FOR2(x,y,z) for((x)=(y);(x)<(z);(x)++)
#define FORD2(x,y,z) for((x)=(y);(x)>=(z);(x)--)
const
int
maxn = 1000 + 10;
int
a[maxn];
int
main()
{
int
n;
while
(~getint2(n) && n){
while
(1){
FOR(i,0,n){
getint2(a[i]);
if
(!a[0])
break
;
}
if
(!a[0])
break
;
int
cur = 1,i = 0;
stack<
int
> s;
while
(i < n){
if
(a[i] == cur){
i++;cur++;
continue
;
}
else
if
(a[i] > cur){
s.push(cur);cur++;
}
else
{
if
(s.top() != a[i]){
puts
(
"No"
);
break
;
}
s.pop();i++;
}
}
if
( i == n)
puts
(
"Yes"
);
}
puts
(
""
);
}
return
0;
}
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来自为知笔记(Wiz)