hdoj 5477 A Sweet Journey （二分查找）

A Sweet Journey

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 729    Accepted Submission(s): 377

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

12 2 2 51 23 4

 

Sample Output

Case #1: 0

 

Source

2015 ACM/ICPC Asia Regional Shanghai Online

慢慢的，对二分有了认识，二分就是先确定一个最小的可能范围 （但要满足条件），然后进行一分为二，判断中间值是否满足题意，

代码：

#include <iostream>#include <cstdio>using namespace std;int t,k,n,a,b,l;int s[100010],e[100010];int judge(int sum)//判断中间值是否可行{    int per=0;    for(int i=1;i<=n;i++)    {        sum+=(s[i]-per)*b;        sum-=(e[i]-s[i])*a;        if(sum<0)            return false;        else            per=e[i];    }    return true;}int main(){    k=1;    int left,right,mid,ans;    scanf("%d",&t);    while(t--)    {        scanf("%d%d%d%d",&n,&a,&b,&l);        for(int i=1;i<=n;i++)        {            scanf("%d%d",&s[i],&e[i]);        }        left=0;        right=l*10+1;//开始脑子短路，写成了2，而a b的值最大为10呢        while(right>=left)        {            mid=(left+right)>>1;            if(judge(mid))            {                ans=mid;                right=mid-1;            }            else                left=mid+1;        }        printf("Case #%d: %d\n",k++,ans);    }    return 0;}