hdu 5477__A Sweet Journey
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A Sweet Journey
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 403 Accepted Submission(s): 223
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 meansRi<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:
Make sure intervals are not overlapped which means
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
12 2 2 51 23 4
Sample Output
Case #1: 0
Source
2015 ACM/ICPC Asia Regional Shanghai Online
#include<cstdio>int main(){ int t,n,l[110],r[110]; scanf("%d",&t); int ans,A,B,L,cas=1; while(t--) { scanf("%d%d%d%d",&n,&A,&B,&L); // A代表cost int now=0,ans=0; for(int i=0;i<n;i++) { scanf("%d%d",&l[i],&r[i]); if(i==0) { if(l[i]>0) { now+=(l[i]-0)*B; } } else { now+=(l[i]-r[i-1])*B; } int cnt=(r[i]-l[i])*A; if(now<cnt) { ans+=cnt-now; now=0; } else { now-=cnt; } } printf("Case #%d: %d\n",cas++,ans); } return 0;}
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