题目链接: OpenJudge - 1684:Dynamic Declaration Language (DDL)

题目描述: 

评价: 一道单纯的大模拟题，算法难度不大，需要非常细心，锻炼我们的耐心和能力！值得一做

分析: 

总结一下题目的条件，这种语言，有2种错误:

1. 变量重复定义

2. 变量使用未声明

这题有一个需要注意的点就是: goto 语句是可以向前跳转的，因此我们需要先把所有的输入存下来，然后再模拟程序的运行过程。(我一开始没考虑到这个，卡了非常久，一直调不出来)

然后别的地方其实没啥难度，只要细心处理就好. 

参考代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
using namespace std;
const int N=110;
int t,n;
int cur_line=0,last_line;
int end_flag=0;
int num_pro;
string s;
string s_ori;
map<string,int> mp;
map<string,int> used;
string guan[10]={"dcl","inc","goto","dec","end"};
struct node{
	string ops;
	string x;
	int value;
};
node arr[N];
string change(string g){
	for(int i=0;i<g.size();i++){
		if(g[i]>='A'&&g[i]<='Z'){
			g[i]-=('A'-'a');
		}
	}
	return g;
}
int check_num(string g){
	int flag=1;
	for(int i=0;i<g.size();i++){
		if(g[i]>='0'&&g[i]<='9'){
			continue;
		}
		else return 0;
	}
	return 1;
}
int change_num(string g){
	int num=0;
	for(int i=0;i<g.size();i++){
		num=num*10+(g[i]-'0');
	}
	return num;
}
int main(void){
	
	scanf("%d",&t);
	while(t--){
		scanf("%d",&n);
		num_pro++;
		printf("%d\n",num_pro);
		last_line=0;
		cur_line=0;
		end_flag=0;
		mp.clear();
		used.clear();
		for(int i=1;i<=n;i++){
			cin >> s;	
			s_ori=s;
			s=change(s);
			if(s==guan[0]){ //dcl
				string id;
				cin >> id;
				arr[i]=node{"dcl",id,0};
			}
			else if(s==guan[1]){ // inc x
				string id;
				cin >> id;
				arr[i]=node{"inc",id,0};
			}
			else if(s==guan[2]){ // goto i  or goto id i
				string line_num;
				string id;
				cin >> id;
				if(check_num(id)){ // goto i
					int num=change_num(id);
					arr[i]=node{"goto","-",num};
				}
				else{ // goto id i 
					int num;
					cin >> num;
					arr[i]=node{"goto",id,num};
				}
			}
			else if(s==guan[3]){ // dec x
				string id;
				cin >> id;
				arr[i]=node{"dec",id,0};

			}
			else if(s==guan[4]){ //end
				arr[i]=node{"end","-",0};
			}
			else{  // id = ic; where ic is constant
				string id;
				id=s_ori;
				string op;
				cin >> op;
				int num;
				cin >> num; 
				arr[i]=node{id,id,num};
			}
		}
		cur_line=1;
		while(cur_line<=n){
			s=arr[cur_line].ops;
			if(s==guan[0]){ //dcl
				string id;
				id=arr[cur_line].x;
				if(mp.count(id)&&used[id]==0){
					printf("%d %d\n",cur_line,1);
					cur_line++;
					continue;
				}
				cur_line++;
				mp[id]=0; 
				used[id]=0; // indicate that the x has not been used.
			}
			else if(s==guan[1]){ // inc x
				string id;
				id=arr[cur_line].x;
				if(!mp.count(id)){
					printf("%d %d\n",cur_line,2);
					cur_line++;
					continue;
				}
				cur_line++;
				used[id]=1;// id has been used
				mp[id]++;
			}
			else if(s==guan[2]){ // goto i  or goto id i
				string line_num;
				string id;
				id=arr[cur_line].x;
				if(id=="-"){ // goto i
					int num=arr[cur_line].value;
					cur_line=num;
					continue;
				}
				else{ // goto id i 
					int num;
					num=arr[cur_line].value;
					if(!mp.count(id)){
						printf("%d %d\n",cur_line,2);
						cur_line++;
						continue;
					}
					used[id]=1;
					if(mp[id]>0){
						cur_line=num;
						continue;
					}
					cur_line++;
				}
			}
			else if(s==guan[3]){ // dec x
				string id;
				id=arr[cur_line].x;
				if(!mp.count(id)){
					printf("%d %d\n",cur_line,2);
					cur_line++;
					continue;
				}
				cur_line++;
				used[id]=1;// id has been used
				mp[id]--;
			}
			else if(s==guan[4]){ //end
				break;
			}
			else{  // id = ic; where ic is constant
				string id;
				id=arr[cur_line].x;
				int num;
				num=arr[cur_line].value;
				if(!mp.count(id)){
					printf("%d %d\n",cur_line,2);
					cur_line++;
					continue;
				}
				mp[id]=num;
				used[id]=1;
				cur_line++;
			}
		}
			
	}	
	
	return 0;
}