思路：水题，标记一下扫一遍就可以了
#include<bits/stdc++.h>
int vis[100005];
int main()
{
int T,cas=1;
scanf("%d",&T);
while(T--)
{
int n,a,b,l;
memset(vis,0,sizeof(vis));
scanf("%d%d%d%d",&n,&a,&b,&l);
for(int i=0;i<n;i++)
{
int li,ri;
scanf("%d%d",&li,&ri);
for(int j=li;j<ri;j++) vis[j]=1;
}
int ans=0,sum=0;
for(int i=0;i<l;i++)
{
if(vis[i]==0) ans+=b;
else ans-=a;
if(ans<0)
{
sum+=ans;
ans=0;
}
}
printf("Case #%d: %d\n",cas++,-sum);
}
return 0;
}
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
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
