題干：

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≤501≤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,RiLi,Ri, which represents the interval?[Li,Ri][Li,Ri]?is swamp.?
1≤n≤100,1≤L≤105,1≤A≤10,1≤B≤10，1≤Li<Ri≤L1≤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+1Ri<Li+1?for each i (1≤i<n1≤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

題目大意：

Mr.D 帶著他的女朋友出去旅行，資金缺乏，就騎著自行車出發(fā)了！路途中會遇到沼澤地和平坦路。每在沼澤地騎行一米，Mr.D能量值減少a,每在平坦大路上騎行一米就恢復(fù)能量值b。為了能成功到達目的地，在出發(fā)前至少需補充多少能量。

解題報告：

跟之前做過的一個機器人的差不多，數(shù)形結(jié)合一下，橫軸是到原點的距離，縱軸是能量變化，先假設(shè)原點的能量是0，最后輸出圖像的最低點就可以了。

AC代碼：

#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<vector> #include<cctype> using namespace std; typedef long long ll; const int maxn=1e4+5; int n,L,a,b; int l[maxn],r[maxn]; int main() {int t,i,j,k,cnt=0,nf;ll sum,ans;cin>>t;r[0]=0;for(;t;t--){scanf("%d%d%d%d",&n,&a,&b,&L); sum=ans=nf=0;if(nf) continue;for(i=1;i<=n;i++){scanf("%d%d",l+i,r+i);sum+=1LL*(l[i]-r[i-1])*b;sum-=1LL*(r[i]-l[i])*a;if(sum<0) {ans+=-sum;sum=0;}}printf("Case #%d: %lld\n",++cnt,ans);}return 0; }

?

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