2015 ACM/ICPC Asia Regional Shanghai Online 1010题

[题目链接]
(http://acm.hdu.edu.cn/contests/contest_showproblem.php?pid=1010&cid=623)
题目描述:
A Sweet Journey

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2619 Accepted Submission(s): 1007

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)
2015 ACM/ICPC Asia Regional Shanghai Online 1010题_第1张图片

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

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

bool flag[100010];

int main()
{
    int T,n,a,b,R;
    int l,r;
    cin>>T;
    int k=1;
    while(T--)
    {
        scanf("%d%d%d%d",&n,&a,&b,&R);
        memset(flag,false,sizeof(flag));
        while(n--)
        {
            scanf("%d%d",&l,&r);
            for(int i=l;i<=r-1;i++)
            {
                flag[i]=true;
            }
        }
        int ans,min;
        ans=0;
        min=0xfffffff;
        for(int  i=0;i<R;i++)
        {
            if(flag[i]==false)
            {
                ans+=b;
            }
            else if(flag[i]==true)
            {
                ans-=a;
            }
            if(ans<min)
            {
                min=ans;
            }
        }
        if(min>=0)
        printf("Case #%d: 0\n",k++);//特别注意,min>=0时,输出0即可
        else
        {
            printf("Case #%d: %d\n",k++,-min);
        }
    }
    return 0;
}

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