【杭电oj】5477 - A Sweet Journey（水）

最新推荐文章于 2024-09-22 09:55:28 发布

wyg1997

最新推荐文章于 2024-09-22 09:55:28 发布

阅读量417

点赞数

CC 4.0 BY-SA版权

分类专栏：水题

本文链接：https://blog.youkuaiyun.com/wyg1997/article/details/51849500

水题专栏收录该内容

93 篇文章

订阅专栏

本题描述了一次由MasterDi带领女友进行的自行车旅行挑战，路径由沼泽与平地构成，需计算出发前所需的最小体力值。通过遍历路径，记录遇到沼泽时的体力消耗，最终给出初始所需体力。

点击打开题目

A Sweet Journey

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 773 Accepted Submission(s): 405

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

Sample Output

  
   Case #1: 0

Source

2015 ACM/ICPC Asia Regional Shanghai Online

这个专题里比较水的题了，并没有用上二分法。

直接从前往后走就行了，遇见沼泽更新 ans ，最后输出 ans 的相反数就行了。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std; 
int main()
{
	int u;
	int n,A,B,l;
	int mapp[100000+11];		//0表示平原，1表示沼泽，表示该点与下一点之间的地形 
	int s,ans;
	int ant = 1;
	scanf ("%d",&u);
	while (u--)
	{
		scanf ("%d %d %d %d",&n,&A,&B,&l);
		memset (mapp,0,sizeof (mapp));
		while (n--)
		{
			int l,r;
			scanf ("%d %d",&l,&r);
			for ( ; l < r ; l++)
				mapp[l] = 1;
		}
		printf ("Case #%d: ",ant++);
		ans = 0;
		s = 0;		//当前体力 
		for (int i = 0 ; i < l ; i++)
		{
			if (mapp[i])		//如果该点是沼泽，则耗费体力并更新ans
			{
				s -= A;
				ans = min (ans , s);
			}
			else
				s += B;
		}
		printf ("%d\n",-ans);
	}
	return 0;
}