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 (1t50), 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.
1n100,1L105,1A10,1B101Li<RiL.
Make sure intervals are not overlapped which means Ri<Li+1 for each i (1i<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
 
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;
}

 

posted @ 2019-07-24 09:42  明霞  阅读(121)  评论(0编辑  收藏  举报