J - A Sweet Journey
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)
InputIn 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.
OutputFor 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 4Sample Output
Case #1: 0
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int n,a,b,l;
struct node
{
int l,r;
}p[105];
bool cmp(node x,node y)
{
return x.l<y.l;
}
int main()
{
int m;
cin>>m;
int cnt=1;
while(m--)
{
cin>>n>>a>>b>>l;
for(int i=1;i<=n;i++)
{
cin>>p[i].l>>p[i].r;
}
sort(p+1,p+n+1,cmp);//以防沼泽输入时，顺序不对
int e=0,sum=0,res=0;
for(int i=1;i<=n;i++)
{
sum+=(p[i].l-e)*b;
int num=p[i].r-p[i].l;
if(sum/a>=num)
{
sum-=num*a;
}
else
{
res+=num*a-sum;
sum=0;
}
e=p[i].r;
}
printf("Case #%d: %d\n",cnt++,res);
}
return 0;
}