HDU 5477: A Sweet Journey
A Sweet Journey
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 25 Accepted Submission(s): 12
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
1 2 2 2 5 1 2 3 4
Sample Output
Case #1: 0
只能切一切这种菜题了。。。
题意就是一个人去旅行,有沼泽地有平坦地,平坦地可以涨力气,沼泽地耗力气。问如果能顺利到达目的地的话,一开始要准备多少力气。
代码:
#include <iostream> #include <algorithm> #include <cmath> #include <vector> #include <string> #include <cstring> #pragma warning(disable:4996) using namespace std; int dis[100005]; int main() { int test,i,j,n,A,B,L,temp1,temp2,ans; scanf("%d",&test); for(i=1;i<=test;i++) { memset(dis,0,sizeof(dis)); scanf("%d%d%d%d",&n,&A,&B,&L); for(j=1;j<=n;j++) { scanf("%d%d",&temp1,&temp2); dis[temp2]=(-1)*(temp2-temp1)*(A+B); } ans=0; for(j=1;j<=L;j++) { dis[j]=dis[j-1]+dis[j]+B; ans=min(ans,dis[j]); } printf("Case #%d: %d\n",i,-1*ans); } return 0; }
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