[NOIP2017 TG D1T2]时间复杂度

题目大意：略

题解：模拟

卡点：1.数组忘清空

（考场代码风格独特。。。）

C++ Code：

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int inf=2000;
int T,n,F,E,top,ls,ans,temp,can,w;
char ch,s[10000];
string p;
bool pd,halt,llw[10000];
bool read(int &x){
    char t=getchar();bool pd=0;
    while(t<'0'||t>'9'){
        if (t=='^')pd=1;
        t=getchar();
    }
    for (x=t^48,t=getchar();t>='0'&&t<='9';t=getchar())x=x*10+(t^48);
    return pd;
}
int main(){
//	freopen("complexity.in","r",stdin);
//	freopen("complexity.out","w",stdout);
    scanf("%d",&T);
    while (T--){
        scanf("%d",&n);
//		printf("Debug:%d\n",n);
        pd=read(w);
//		printf("D:%d\n",w);
        getline(cin,p);
        top=0;can=inf;
        halt=false;
        ans=temp=0;
        for (int i=0;i<n;i++){
            int x,y;
            char cha;
            getline(cin,p);
            ls=p.length();
//			cout << "lalal" << p << endl;
            if (halt)continue;
            if (p[0]=='F'){
                int now;
                for (now=1;now<ls;now++)if (p[now]!=' ')break;
                ch=p[now];
                for (int j=0;j<top;j++)if (ch==s[j]){
                    puts("ERR");
                    halt=1;
                    continue;
                }
                if (halt)continue;
                s[top++]=ch;
                llw[top-1]=0;
                if (can<top)continue;
                for (now++;now<ls;now++)if (p[now]!=' ')break;
                if (p[now]=='n')x=inf;
                    else {
                        for (x=p[now]^48,now++;p[now]>='0'&&p[now]<='9';now++)x=x*10+(p[now]^48);
                    }
                for (now++;now<ls;now++)if (p[now]!=' ')break;
                if (p[now]=='n')y=inf;
                    else {
                        for (y=p[now]^48,now++;p[now]>='0'&&p[now]<='9';now++)y=y*10+(p[now]^48);
                    }
//				printf("DEBUG:%d %d\n",x,y);
                if (x<y&&y==inf)temp++,llw[top-1]=1;
                if (temp>ans)ans=temp;
//                printf("%d\n", temp);
                if (x==y)llw[top-1]=0;
                if (x<y&&y!=inf)llw[top-1]=0;
                if (x>y)can=top,llw[top-1]=0;
            }else{
                if (p[0]=='E'){
                    top--;
                    if (top<0){
                        puts("ERR");
                        halt=1;
                        continue;
                    }
                    if (llw[top])temp--;
                    if(can>=top)can=inf;
                }
            }
        }
        if (top&&!halt){
            puts("ERR");
            halt=1;
        }
        if (!halt){
            if (pd){
                if (ans==w)puts("Yes");
                    else puts("No");
            }else if (ans==0)puts("Yes");
                else puts("No");
        }
    }
    return 0;
}