怒刷DP之 HDU 1260
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
System Crawler (2015-09-06)
Description
A good approach, reducing the total time of tickets selling, is let adjacent people buy tickets together. As the restriction of the Ticket Seller Machine, Joe can sell a single ticket or two adjacent tickets at a time.
Since you are the great JESUS, you know exactly how much time needed for every person to buy a single ticket or two tickets for him/her. Could you so kind to tell poor Joe at what time could he go back home as early as possible? If so, I guess Joe would full of appreciation for your help.
Input
1) An integer K(1<=K<=2000) representing the total number of people;
2) K integer numbers(0s<=Si<=25s) representing the time consumed to buy a ticket for each person;
3) (K-1) integer numbers(0s<=Di<=50s) representing the time needed for two adjacent people to buy two tickets together.
Output
Sample Input
2
20 25
40
1
8
Sample Output
08:00:08 am
#include <iostream>
#include <cstdio>
#include <string>
#include <queue>
#include <vector>
#include <map>
#include <algorithm>
#include <cstring>
#include <cctype>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <climits>
using namespace std; const int SIZE = ;
int DP[SIZE],S[SIZE],H[SIZE]; int main(void)
{
int t,n; scanf("%d",&t);
while(t --)
{
memset(DP,,sizeof(DP));
scanf("%d",&n);
for(int i = ;i <= n;i ++)
scanf("%d",&S[i]);
for(int i = ;i <= n - ;i ++)
scanf("%d",&H[i]); if(n == )
printf("08:00:%02d am\n",S[]);
else
{
for(int i = ;i <= n;i ++)
DP[i] = min(DP[i - ] + S[i],DP[i - ] + H[i - ]);
int sec = DP[n];
int hour = sec / ( * );
sec -= hour * * ;
int min = sec / ;
sec -= min * ;
hour += ; bool flag_am = true;
if(hour >= )
{
flag_am = false;
hour -= ;
}
printf("%02d:%02d:%02d %s\n",hour,min,sec,flag_am ? "am" : "pm");
}
} return ;
}
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