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  • HDU 5477: A Sweet Journey

    A Sweet Journey

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
    Total Submission(s): 25    Accepted Submission(s): 12


    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
    最近状态一直不怎么好,想了很多问题都没有想出答案。当然了,我这种状态好了也不会好到哪里去。。。

    只能切一切这种菜题了。。。

    题意就是一个人去旅行,有沼泽地有平坦地,平坦地可以涨力气,沼泽地耗力气。问如果能顺利到达目的地的话,一开始要准备多少力气。

    代码:

    #include <iostream>  
    #include <algorithm>  
    #include <cmath>  
    #include <vector>  
    #include <string>  
    #include <cstring>  
    #pragma warning(disable:4996)  
    using namespace std;
    
    int dis[100005];
    
    int main()
    {
    	int test,i,j,n,A,B,L,temp1,temp2,ans;
    	scanf("%d",&test);
    	for(i=1;i<=test;i++)
    	{
    		memset(dis,0,sizeof(dis));
    		scanf("%d%d%d%d",&n,&A,&B,&L);
    		for(j=1;j<=n;j++)
    		{
    			scanf("%d%d",&temp1,&temp2);
    			dis[temp2]=(-1)*(temp2-temp1)*(A+B);
    		}
    		ans=0;
    		for(j=1;j<=L;j++)
    		{
    			dis[j]=dis[j-1]+dis[j]+B;
    			ans=min(ans,dis[j]);
    		}
    		printf("Case #%d: %d
    ",i,-1*ans);
    	}
    	return 0;
    }



    版权声明:本文为博主原创文章,未经博主允许不得转载。

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  • 原文地址:https://www.cnblogs.com/lightspeedsmallson/p/4899562.html
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