ACM--过沼泽--模拟--HDOJ 5477--A Sweet Journey
2016-05-28 17:13
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HDOJ题目地址:传送门
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 759 Accepted Submission(s): 397
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)
![](http://acm.hdu.edu.cn/data/images/C623-1010-1.jpg)
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.
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.
Sample Input
Sample Output
题意:一个人去旅行,路上有沼泽和平路,当走平路是增加a点体力,当走沼泽时消耗b点体力,求最开始最少要携带多少点体力
A Sweet Journey
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 759 Accepted Submission(s): 397
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)
![](http://acm.hdu.edu.cn/data/images/C623-1010-1.jpg)
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.
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.
Sample Input
1 2 2 2 5 1 2 3 4
Sample Output
Case #1: 0
题意:一个人去旅行,路上有沼泽和平路,当走平路是增加a点体力,当走沼泽时消耗b点体力,求最开始最少要携带多少点体力
#include<iostream> #include<stdio.h> #include<memory.h> #include<algorithm> using namespace std; struct Node{ int begin; int end; int chazhi; }zhaoze[101]; bool cmp(Node a,Node b){ if(a.begin<b.begin) return true; return false; } int main(){ int n,m,i,a,b,l,kaishi,jieshu,result,temp,index=1; cin>>n; while(n--){ cin>>m>>a>>b>>l; for(i=0;i<m;i++){ cin>>zhaoze[i].begin>>zhaoze[i].end; zhaoze[i].chazhi=zhaoze[i].end-zhaoze[i].begin; } sort(zhaoze,zhaoze+m,cmp); result=0; temp=0; kaishi=0; jieshu=0; for(i=0;i<m;i++){ temp=temp+(zhaoze[i].begin-jieshu)*b-zhaoze[i].chazhi*a; jieshu=zhaoze[i].end; if(result>temp){ result=temp; } } temp+=(l-jieshu)*b; if(result>temp){ result=temp; } if(result<0){ printf("Case #%d: %d\n",index,-result); }else{ printf("Case #%d: 0\n",index); } index++; } }
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