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其实很简单的!
把过道分成200份,定义一个整型数组统计每块过道上搬运经过的次数,
最后排序,输出最大的次数乘以10即可!
如果还不明白,看一下代码吧!A过的!
#include<iostream>
#include<algorithm>
using namespace std;
int main()
{ int n,num,a[200][2],b[200];
cin>>n;
while(n--)
{cin>>num;
for(int m=0;m<200;m++)
b[m]=0;
for(int i=0;i<num;i++)
{cin>>a[i][0]>>a[i][1];
if(a[i][0]>a[i][1])
{int temp=a[i][0];a[i][0]=a[i][1];a[i][1]=temp;}
}
for(int j=1;j<=200;j++)
for(int k=0;k<num;k++)
if((((a[k][0])<=2*j-1)&&((a[k][1])>=2*j-1)||(((a[k][0])<=2*j)&&((a[k][1])>=2*j))))
b[j-1]++;
sort(&b[0],&b[200]);
if(b[199]!=0) cout<<b[199]*10<<endl;
else cout<<'10'<<endl;
}
return 0;
}
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