0526 HDU#5477&G2n#A-A Sweet Journey

摘要:
按照一定的消耗方式，如何最少的使用能量。
原题目链接：HDU
- 5477

 Sweet
Journey

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) 

InputIn the first line there is an integer t (1≤t≤501≤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,RiLi,Ri,
which represents the interval [Li,Ri][Li,Ri] is
swamp. 
1≤n≤100,1≤L≤105,1≤A≤10,1≤B≤10，1≤Li<Ri≤L1≤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+1Ri<Li+1 for
each i (1≤i<n1≤i<n). 

Others are all flats except the swamps. 

OutputFor 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

来源： https://cn.vjudge.net/problem/description/56056?1495774615000

题目认识：
假定开始时已经是恰好最优状态 并设为状态0，依照消耗方式进行模拟。最后看状态的值是多少即可确定开始时的最低是多少。

注意：
计算需要补给的是多少。其他的不用改变状态值。

日期：
2017 5 26
代码：

#include <cstdio>

#include <algorithm>

#define MAX 0

int getans(){

int n,A,B,L;

int need=0,s=0;//strength

int li,ri,lastRi=0;

scanf("%d%d%d%d",&n,&A,&B,&L);

while(n--){

scanf("%d%d",&li,&ri);

s+=B*(li-lastRi);

s-=A*(ri-li);

if(s<0){need+=s;s=0;}

lastRi=ri;

}

return -need;

}

int main(){

int t;

scanf("%d",&t);

for(int i=1;i<=t;i++){

printf("Case #%d: %d\n",i,getans());

}

return 0;

}