CCF computer vocational qualification certification exam 201809-2 Food Buyers

The following content is too naive ..

Description:

Time limit:
1.0 s
Memory limit:
256.0 MB
Problem description:
Problem description
H and W came to a street, And they bought food separately. They could describe the process of buying food. They went to the store to buy some food and then loaded the food into a square next to them, both of them need to buy n kinds of food, so they also need to install n cars. Specifically, for small h, there are N non-Intersecting time periods [A1, B1], [A2, B2]... [An, BN] when loading a car, for small W, there are n different time periods [C1, D1], [C2, D2]... [CN, DN] loading. [S, T] indicates the period from time s to time t, and the length of time is t-s.
Because they are good friends, they will chat when they are loading cars in the square. They want to know how long they can chat.
Input Format
The first line of the input contains a positive integer N, indicating the number of time periods.
In the next n rows, there are two pieces of AI and Bi in each row, which describe the loading time periods of small h.
The next n rows have two CI, Di, and describe the loading time periods of W.
Output Format
Output a row, a positive integer, indicating how long the two can chat.
Sample Input
4
1 3
5 6
9 13
14 15
2 4
5 7
10 11
13 14
Sample output
3
Data scale and conventions
For all evaluation cases, 1 ≤ n ≤ 2000, AI <Bi <AI + 1, CI <di <Ci + 1, for all I (1 ≤ I ≤ n) yes, 1 ≤ AI, Bi, CI, di ≤ 1000000.

 

1. Simple Solution, applicable to all situations

 1 #include<iostream> 2 using namespace std; 3 int main(){ 4     int n; 5     cin>>n; 6     int a[n],b[n],c[n],d[n]; 7     for(int i=0;i<n;i++){ 8         cin>>a[i]>>b[i]; 9     }10     for(int i=0;i<n;i++){11         cin>>c[i]>>d[i];12     }13     int count=0;14     int i=0,j=0;15     while(i<n&&j<n){16         if(a[i]<=c[j]){17             if(b[i]<=d[j]&&b[i]>c[j]) {18                 count+=b[i]-c[j];i++;19             }20             else {21                 if(b[i]<=c[j]) i++;22                 else{23                     count+=d[j]-c[j];j++;24                 }    25             }26                 27         }28         else{29             if(b[i]>=d[j]&&d[j]>a[i]){30                 count+=d[j]-a[i];j++;31             }32             else{33                 if(d[j]<=a[i]) j++;34                 else{35                     count+=b[i]-a[i];i++;36                 }37                 38             }39         }40     41     }        42     cout<<count;43     return 0;44 } 

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2. Summarize the problem

 1 #include<iostream> 2 using namespace std; 3 int main(){ 4     int n; 5     cin>>n; 6     int a[n],b[n],c[n],d[n]; 7     for(int i=0;i<n;i++){ 8         cin>>a[i]>>b[i]; 9     }10     for(int i=0;i<n;i++){11         cin>>c[i]>>d[i];12     }13     int count=0;int x,y;14     for(int i=0;i<n;i++){15         for(int j=0;j<n;j++){16             count+=max(0,min(b[i],d[j])-max(a[i],c[j]));17         }18     }19     cout<<count;20     return 0;21 }

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3. The most understandable and ingenious

 1 #include<iostream> 2 using namespace std; 3 int ojbk[1000000];     4 int main(){ 5     int n; 6     cin>>n; 7     int a[n],b[n],c[n],d[n]; 8     for(int i=0;i<1000000;i++){ 9         ojbk[i]=0;10     }11     for(int i=0;i<n;i++){12         cin>>a[i]>>b[i];13         for(int j=a[i];j<b[i];j++){14             ojbk[j]++;15         }16     }17     for(int i=0;i<n;i++){18         cin>>c[i]>>d[i];19         for(int j=c[i];j<d[i];j++){20             ojbk[j]++;21         }22     }23     int count=0;24     for(int i=0;i<1000000;i++){25         if(ojbk[i]>1) count++;26     }27     cout<<count;28     return 0;29 }

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CCF computer vocational qualification certification exam 201809-2 Food Buyers