HDU 5477 A Sweet Journey（亚洲区水题）

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

本题的大致题意为：一个人想去旅游，问他一开始最小需要的体力是多少。有两种地形，沼泽和平原，如果是沼泽的话，每米消耗A的体力，在平原每米恢复B的体力。要注意的一点是在这个过程中这个人的最小体力值为0，不可以小于0；利用这个条件求出在开始时具有的最小体力值。
本题个人感觉其实并不算太难，应该可以算上一道模拟题，为什么呢，，只要你读懂题之后，这道题就换成了最小值更新的问题，所以呢，，还是好好锻炼自己的思维吧。
下面附上AC代码：

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;

int main()
{
    int t;
    int iCase=0;
    scanf("%d",&t);
    while(t--)
    {
        iCase++;
        int x,y;
        int flag=0;
        int n,a,b,l;
        int sum=0;
        int minn=1000000;
        scanf("%d%d%d%d",&n,&a,&b,&l);
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&x,&y);
            sum=sum+(x-flag)*b;
            sum=sum-(y-x)*a;
            minn=min(minn,sum);
            flag=y;
        }
        if(minn>=0)
        {
            printf("Case #%d: 0\n",iCase);
        }
        else
        {
            printf("Case #%d: %d\n",iCase,-minn);
        }
    }
    return 0;
}