2 2 6 08:00 09:00 5 08:59 09:59 2 6 08:00 09:00 5 09:00 10:00
11 6
注释里的代码 t*n*时间区间=100*10000*1440≈10^9 应该超时的. 居然能过...
#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
int a[2000];
int T,m,i;
scanf("%d",&T);
while(T--)
{
int n,h1,h2,m1,m2;
int sum1,sum2,maxx=0;
scanf("%d",&m);
memset(a,0,sizeof(a));
while(m--)
{
scanf("%d%d:%d%d:%d",&n,&h1,&m1,&h2,&m2);
sum1=h1*60+m1;
sum2=h2*60+m2;
a[sum1]+=n;
a[sum2]-=n;
}
maxx=a[0];
for(i=1;i<2000;i++)
{
a[i]+=a[i-1];
maxx=max(maxx,a[i]);
}
printf("%d\n",maxx);
}
return 0;
}
/*
#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
int a[2000];
int T,m,i;
scanf("%d",&T);
while(T--)
{
int n,h1,h2,m1,m2;
int sum1,sum2,maxx=0;
scanf("%d",&m);
memset(a,0,sizeof(a));
while(m--)
{
scanf("%d%d:%d%d:%d",&n,&h1,&m1,&h2,&m2);
sum1=h1*60+m1;
sum2=h2*60+m2;
for(i=sum1+1;i<=sum2;i++)
{
a[i]+=n;
}
}
for(i=0;i<2000;i++)
{
maxx=max(maxx,a[i]); //找出最大值
}
printf("%d\n",maxx);
}
return 0;
}
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原文地址：http://blog.csdn.net/u013532224/article/details/38184783
