标签：
A Sweet Journey
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 741 Accepted Submission(s): 383
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
Sample Output
Source
题意：在区间 [0, L] 中有沼泽和陆地，经过沼泽每米损失a体力，经过陆地每米回复b体力，要从0到L；求在0处所需的最小体力。
思路：只需要求经过一块沼泽需要的最大体力就行
#include <stdio.h>
#include <algorithm>
using namespace std;
int main ()
{
int T,n,a,b,L;
scanf("%d",&T);
for (int i=1; i<=T; i++)
{
scanf("%d%d%d%d",&n,&a,&b,&L);
int p=0,sum=0,minx=99999;//p用来标记上一个沼泽结束位置
int l,r;
while (n--)
{
scanf("%d%d",&l,&r);
sum = sum+b*(l-p)-a*(r-l);//每经过一片沼泽所剩于体力
minx = min(minx, sum);//途中所需的最大体力
p = r;//更新p
}
printf("Case #%d: ",i);
if (minx < 0)
printf("%d\n",-minx);
else
printf("0\n");
}
return 0;
}
HDU 5477 A Sweet Journey
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原文地址：http://blog.csdn.net/yao1373446012/article/details/51355001
