NOIP 2017 时间复杂度 (模拟)

题目大意：略

傻了吧唧的我wa了好几次

我的模拟功底真的不咋地

linux下用gets会报错，我用的fgets

 1 #include <string>
 2 #include "stdio.h"
 3 #include <cstring>
 4 #include <iostream>
 5 #include <algorithm>
 6 #define ll long long 
 7 #define N 1010
 8 #define rint register int
 9 #define ll long long 
10 #define il inline
11 #define idx(x) (x-'a'+1)
12 #define idg(x) (x-'0')
13 using namespace std;
14 //re
15 int T,n,cnt,j;
16 int use[30];
17 char str[110][20];
18 inline bool isdig(char c)
19 {return ('0'<=c&&c<='9')?1:0;}
20 int dfs()
21 {
22     int ans=0,now=j,val=0,cal=0;
23     if(str[j][0]=='E') return -1;
24     if(str[j][0]=='F')
25     {
26         if(use[idx(str[j][2])])
27             return -1;
28         use[idx(str[j][2])]=1;
29         while(str[j+1][0]=='F')
30         {
31             j++,cal=dfs();
32             if(cal==-1) return -1;
33             ans=max(ans,cal);
34         }  
35         int k=4;
36         while(isdig(str[now][k]))
37             val=val*10+idg(str[now][k]),k++; 
38         k++;
39         if(str[now][4]=='n'){
40             if(str[now][6]!='n') ans=0;
41         }else if(!isdig(str[now][k])){
42             ans++;
43         }else{
44             int w=0;
45             while(isdig(str[now][k]))
46                 w=w*10+idg(str[now][k]),k++;
47             if(val>w) ans=0;
48         }
49     }
50     use[idx(str[now][2])]=0;
51     j++;
52     if(str[j][0]!='E') return -1;
53     else return ans;
54 }
55 
56 int main()
57 {
58     //freopen("complexity6.in","r",stdin);
59     scanf("%d",&T);
60     for(int t=1;t<=T;t++)
61     {
62         int q=0;
63         scanf("%d",&n);
64         memset(str,0,sizeof(str));
65         memset(use,0,sizeof(use));
66         scanf("%s\n",str[0]);
67         for(int i=1;i<=n;i++)
68             fgets(str[i],20,stdin);
69         int ret=0,ans=0;j=1;
70         for(;j<=n;j++){
71             ans=dfs();
72             if(ans==-1){ret=-1;printf("ERR\n");break;}
73             else {ret=max(ret,ans);}
74         }
75         if(ret==-1) continue;
76         int k=0,flag=0;
77         while(1){
78             if(str[0][k]=='n') flag=1;
79             if(!isdig(str[0][k]))
80                 {k++;continue;}
81             while(isdig(str[0][k]))
82                 q=q*10+idg(str[0][k]),k++;
83             break;
84         }
85         if(flag==0) q=0;
86         if(ret==q) printf("Yes\n");
87         else printf("No\n");
88     }
89     return 0;
90 }

 

posted @ 2018-10-01 09:56  guapisolo  阅读(186)  评论(0)    收藏  举报