把No写成NO,WA了一发……
现在看这题也不难……
用一个栈,记一下前面F的字母,是否合法,合法的有多长,每次入栈弹栈即可
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int N=105;
int T,n,top,co[305],mx,fl,er,va;
char o[10],c[N];
struct qwe
{
char c;
int fl,va;
}s[N];
int main()
{
scanf("%d",&T);
while(T--)
{
scanf("%d%s\n",&n,o);//cerr<<" "<<n<<endl;
top=0,mx=0,fl=0,er=0,va=0;
memset(co,0,sizeof(co));
for(int i=1;i<=n;i++)
{
gets(c);//cerr<<"???"<<i<<" "<<c<<endl;
if(c[0]=='E')
{
if(!top)
er=1;
else
{
co[s[top].c]--;
fl-=s[top].fl;
va-=s[top].va;
top--;
}
}
else
{
if(co[c[2]])
er=1;
s[++top].c=c[2];
co[c[2]]++;
int w=4,x=0,fx=-1,y=0,fy=-1;
while(c[w]>='0'&&c[w]<='9')
fx=1,x=x*10+c[w]-48,w++;
w=max(w+1,6);
while(c[w]>='0'&&c[w]<='9')
fy=1,y=y*10+c[w]-48,w++;
// cerr<<fx<<" "<<x<<" "<<fy<<" "<<y<<endl;
if((fx==-1&&fy!=-1)||(fx!=-1&&fy!=-1&&x>y))
s[top].fl=1,fl++;
else
s[top].fl=0;
if(fx!=-1&&fy==-1)
s[top].va=1,va++;
else
s[top].va=0;
if(!fl)
mx=max(mx,va);
}
}
if(top)
er=1;
int x=0,w=4;
while(o[w]>='0'&&o[w]<='9')
x=x*10+o[w]-48,w++;
// cerr<<x<<" "<<mx<<endl;
if(er)
puts("ERR");
else if((o[2]=='n'&&mx==0)||(o[2]=='n'&&mx!=x))
puts("No");
else
puts("Yes");
}
return 0;
}