HDU 5477 A Sweet Journey(本场的最水题，过程处理好是关键)——2015 ACM/ICPC Asia Regional Shanghai Online

A Sweet Journey

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 47    Accepted Submission(s): 23

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

12 2 2 51 23 4

 

Sample Output

Case #1: 0

 

Source

2015 ACM/ICPC Asia Regional Shanghai Online

 

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题意：一条长L的路，区间为[0,L]，这条路上有陆地和沼泽。当在陆地上行走，每走过x的距离会恢复B*x的体力；当在沼泽上行走时，每走过x的距离会消耗A*x的体力，问一开始至少需要多少体力才能通过这条长L的路。

其实这题只要一个个沼泽处理一遍就可以了，毕竟题目给出的沼泽是不会重叠，而且是递增的(Ri<Li+1)

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<stdio.h>#include<string.h>#include<stdlib.h>#include<queue>#include<stack>#include<math.h>#include<vector>#include<map>#include<set>#include<stdlib.h>#include<cmath>#include<string>#include<algorithm>#include<iostream>#define exp 1e-10using namespace std;const int N = 105;const int inf = 1000000000;const int mod = 2009;int l[N],r[N];int main(){    int t,n,A,B,L,i,k,Max,m,p=1;    scanf("%d",&t);    while(t--)    {        scanf("%d%d%d%d",&n,&A,&B,&L);        for(m=Max=i=0;i<n;i++)            scanf("%d%d",&l[i],&r[i]);        for(k=i=0;i<n;i++)        {            m=m+B*(l[i]-k)-A*(r[i]-l[i]);//当前体力值            Max=max(Max,-m);            k=r[i];        }        printf("Case #%d: %d\n",p++,Max);    }    return 0;}

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