HDU 5477 A Sweet Journey(亚洲区水题)
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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 (
For each test case:
The first line contains four integers, n, A, B, L.
Next n lines, each line contains two integers:
Make sure intervals are not overlapped which means
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;}
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