HDU - 5477 A Sweet Journey

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

 

题意：要我们找需要消耗体力最多的那个地方（即你走一段沼泽，可能会体力为0，这就是你开头需要准备的体力。或是你上一段体力补充完了，这一段又被沼泽消耗光了体力，这也是要补充的体力）。读不懂题真要命

#include <cstdio>
#include <iostream>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <map>
using namespace std;
#define ll long long
const int maxn = 5100;

int t, n, A, B, L, l, r;

int main()
{
    scanf("%d", &t);
    int miao = t;
    while(t--)
    {
        scanf("%d%d%d%d", &n, &A, &B, &L);
        int legacy = 0, position = 0, need = 0;//legacy是剩余耐力，need是需求，position是当前位置
        for(int i = 0; i<n; i++)
        {
            scanf("%d%d", &l, &r);
            legacy += (l-position)*B-(r-l)*A;//（上一段平底回复的）-（这一段澡泽消耗的）
            position = r;//把上一段澡泽的最后一个位置标记下来
            if(legacy<0)//耐力不够时需要额外添加
            {
                need -= legacy;//这时候legacy是负的，所以要减
                legacy = 0;
            }
        }
        printf("Case #%d: %d\n", miao-t, need);
    }
    return 0;
}