hdu 1260 Tickets

1.题目

http://acm.hdu.edu.cn/showproblem.php?pid=1260

2.分析

简答DP，类似于走阶梯(走一步还是两步到达终点)的问题，状态转移方为：f[i]=min(f[i-1]+t1[i],f[i-2]+t2[i]);

决策是：前一个、前两个

算出秒数之后利用时间函数转换成具体的时分秒输出即可，


3.复杂度

4.涉及内容

动态规划

5.感想

6.代码

#include <iostream>
#include <time.h>
#include <string.h>
using namespace std;
unsigned int time1[2001];
long time2[2001];
int min(int a,int b)
{
return a>b?b:a;
}
void converttime(long time,struct tm &t)
{
int hour=0,minute=0,second=0;
second=time%60;
minute=((time-second)/60)%60;	minute%=60;
hour=(time-second-minute*60)/3600;	hour%=24;
t.tm_hour+=hour;	t.tm_min+=minute;	t.tm_sec+=second;
}
int main()
{
freopen("in.txt","r",stdin);
struct tm t;
t.tm_year=2013-1900;	t.tm_mon=4;	t.tm_mday=2;
t.tm_hour=8;	t.tm_min=0;	t.tm_sec=0;	t.tm_isdst=0;
char str[13];
int N,K;
cin>>N;
while(N--)
{
memset(time1,0,sizeof(time1));
memset(time2,0,sizeof(time2));
t.tm_hour=8;	t.tm_min=0;	t.tm_sec=0;
cin>>K;
for(int i=1;i<=K;++i)
cin>>time1[i];
for(int i=1;i<=K-1;++i)
cin>>time2[i+1];
for(int i=2;i<=K;++i)
{
time1[i]=min(time1[i-1]+time1[i],time1[i-2]+time2[i]);
}
converttime(time1[K],t);
strftime(str,13,"%H:%M:%S %p",&t);
cout<<strlwr(str)<<endl;
}
return 0;
}

7.参考文献