Bad practice:
Not difficult to find, you can use the tree recursive solution, error-prone at this point to say something.
1. The use of storage deque several cycles.
On the tail, if was empty, ERR is 2. read E. Note that each mark next cycle should be cleared.
3. When the x, y is n, he is o1, you can search down.
#include<map>
#include<queue>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<deque>
using namespace std;
int t;
int l;
char fzd[10];
int num;
bool flag;
char b[210];
map<char,bool> mp;
deque<int> q;
int tot,pre[210],last[210],other[210],w[210];
void add(int x,int y)
{
tot++;
pre[tot]=last[x];
last[x]=tot;
other[tot]=y;
}
int dfs(int x)
{
int tmp=w[x],sum=0;
for(int p=last[x];p;p=pre[p])
{
int v=other[p];
if(w[v]==-1) continue;
sum=max(sum,dfs(v));
}
return tmp+sum;
}
int main()
{
scanf("%d",&t);
while(t--)
{
tot=0;
q.clear();
memset(last,0,sizeof(last));
memset(w,0,sizeof(w));
flag=0;
num=0;
mp.clear();
scanf("%d",&l);
scanf(" O(%s)",fzd+1);
int len=strlen(fzd+1);
if(len==1) num=0;
else
{
for(int i=3;i<=len-1;i++)
num*=10,num+=fzd[i]-'0';
}
for(int i=1;i<=l;i++)
{
char s;
cin>>s;
if(s=='F')
{
cin>>b[i];
if(mp[b[i]])
flag=1;
mp[b[i]]=1;
char kai[5],ed[5];
scanf("%s",kai+1);
scanf("%s",ed+1);
int l1=strlen(kai+1);
int l2=strlen(ed+1);
int num1=0,num2=0;
for(int j=1;j<=l1;j++)
{
if(kai[j]=='n')
{
w[i]=-1;
continue;
}
num1*=10;
num1+=kai[j]-'0';
}
for(int j=1;j<=l2;j++)
{
if(ed[j]=='n')
{
if(kai[1]!='n')
w[i]=1;
}
num2*=10;
num2+=ed[j]-'0';
}
if(w[i]==0)
{
if(num1<=num2)
w[i]=0;
else w[i]=-1;
}
if(kai[1]=='n'&&ed[1]=='n')
w[i]=0;
if(q.size()) add(q.back(),i);
if(q.empty()) add(0,i);
q.push_back(i);
}
else
{
if(q.empty())
{
flag=1;
continue;
}
mp[b[q.back()]]=0;
q.pop_back();
}
}
if(flag||q.size())
{
printf("ERR\n");
continue;
}
int ans=dfs(0);
if(ans==num)
printf("Yes\n");
else
printf("No\n");
}
return 0;
}