hdoj 5477 A Sweet Journey

E - A Sweet Journey

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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Status

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 ($1 \leq t \leq 50$), 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: $L_i, R_i$, which represents the interval $[L_i, R_i]$ is swamp.

$1 \leq n \leq 100, 1 \leq L \leq 10^5,1 \leq A \leq 10,1 \leq B \leq 10，1 \leq L_i < R_i \leq L$.

Make sure intervals are not overlapped which means $R_i < L_{i + 1}$ for each i ($1 \leq i < n$).

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

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n,a,b,l;
int shu[101000];
int main()
{
int t;scanf("%d",&t);
for (int ca=1;ca<=t;ca++)
{
int x,y;
scanf("%d%d%d%d",&n,&a,&b,&l);
for (int i=1;i<=l;i++)
shu[i]=b;
for (int i=0;i<n;i++)
{
scanf("%d%d",&x,&y);
for (int j=x+1;j<=y;j++)
shu[j]=-a;
}
int s=0,sum=0;
for (int i=1;i<=l;i++)
{
s+=shu[i];
sum=min(sum,s);
}
printf("Case #%d: %d\n",ca,-1*sum);
}
return 0;
}

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n,a,b,l;
int shu[101000];
int main()
{
int t;scanf("%d",&t);
for (int ca=1;ca<=t;ca++)
{
int x,y;
scanf("%d%d%d%d",&n,&a,&b,&l);
memset(shu,0,sizeof(shu));
for (int i=0;i<n;i++)
{
scanf("%d%d",&x,&y);
shu[y]=-1*(y-x)*(a+b);
}
int s=0,sum=0;
for (int i=1;i<=l;i++)
{
s+=shu[i]+b;
sum=min(sum,s);
}
printf("Case #%d: %d\n",ca,-1*sum);
}
return 0;
}