hdu5477 A Sweet Journey（简单模拟）

Link：http://acm.hdu.edu.cn/showproblem.php?pid=5477

A Sweet Journey

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 117 Accepted Submission(s): 60



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

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

Source

2015 ACM/ICPC Asia Regional Shanghai Online



AC code：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
#include<map>
#include<stack>
#include<vector>
#define LL long long
#define MAXN 1000010
using namespace std;
const  int N=20;
struct node{
	int l;
	int r;
	bool operator < (const node &b)const
	{
		return l<b.l;
	}
}road[MAXN];
LL ans,c;
int main()
{
//	freopen("D:\\in.txt","r",stdin);
	int T,i,j,cas,n,a,b,len,temp;
	scanf("%d",&T);
	for(cas=1;cas<=T;cas++)
	{
		scanf("%d%d%d%d",&n,&a,&b,&len);
		for(i=1;i<=n;i++)
		{
			scanf("%d%d",&road[i].l,&road[i].r);
		}
		ans=0;
		temp=0;
		c=0;
		for(i=1;i<=n;i++)
		{
			c+=b*(road[i].l-temp);
			if(c < (road[i].r-road[i].l)*a)
			{
				ans+=((road[i].r-road[i].l)*a-c);
				c=0;
			}
			else
			{
				c-=(road[i].r-road[i].l)*a;
			}
			temp=road[i].r;
		}
		printf("Case #%d: %I64d\n",cas,ans);
	}
  	return 0;
}