A Sweet Journey
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 975 Accepted Submission(s): 504
Problem Description
Master Di plans to take his girlfriend for a travel by bike. Their journey, which can be seen as a line segment of length L, is a road of swamps and flats. In the swamp, it takes A point strengths per meter for Master Di to ride; In the flats, Master Di will regain B point strengths per meter when riding. Master Di wonders：In the beginning, he needs to prepare how much minimum strengths. (Except riding all the time,Master Di has no other choice)
Input
In the first line there is an integer t (1≤t≤50), indicating the number of test cases.
For each test case:
The first line contains four integers, n, A, B, L.
Next n lines, each line contains two integers: Li,Ri, which represents the interval [Li,Ri] is swamp.
1≤n≤100,1≤L≤105,1≤A≤10,1≤B≤10，1≤Li<Ri≤L.
Make sure intervals are not overlapped which means Ri<Li+1 for each i (1≤i<n).
Others are all flats except the swamps.
For each test case:
The first line contains four integers, n, A, B, L.
Next n lines, each line contains two integers: Li,Ri, which represents the interval [Li,Ri] is swamp.
1≤n≤100,1≤L≤105,1≤A≤10,1≤B≤10，1≤Li<Ri≤L.
Make sure intervals are not overlapped which means Ri<Li+1 for each i (1≤i<n).
Others are all flats except the swamps.
Output
For each text case:
Please output “Case #k: answer”(without quotes) one line, where k means the case number counting from 1, and the answer is his minimum strengths in the beginning.
Please output “Case #k: answer”(without quotes) one line, where k means the case number counting from 1, and the answer is his minimum strengths in the beginning.
Sample Input
1
2 2 2 5
1 2
3 4
Sample Output
Case #1: 0
Source
1 //2016.9.13 2 #include <iostream> 3 #include <cstdio> 4 #define N 105 5 6 using namespace std; 7 8 int main() 9 { 10 int T, kase = 0, ans, n, a, b, len, l[N], r[N]; 11 scanf("%d", &T); 12 while(T--) 13 { 14 int tmp = 0; 15 ans = 0;//假设最初需要0点能量 16 scanf("%d%d%d%d", &n, &a, &b, &len); 17 l[0] = r[0] = 0; 18 for(int i = 1; i <= n; i++) 19 { 20 scanf("%d%d", &l[i], &r[i]); 21 tmp = tmp+b*(l[i]-r[i-1])-a*(r[i]-l[i]); 22 if(ans > tmp)ans = tmp; 23 } 24 printf("Case #%d: %d ", ++kase, -ans); 25 } 26 27 return 0; 28 }