H - Tickets HDU 1260 （动态规划）

H - Tickets
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Submit Status Practice HDU 1260

Description
Jesus, what a great movie! Thousands of people are rushing to the cinema. However, this is really a tuff time for Joe who sells the film tickets. He is wandering when could he go back home as early as possible.
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
There are N(1<=N<=10) different scenarios, each scenario consists of 3 lines:
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
For every scenario, please tell Joe at what time could he go back home as early as possible. Every day Joe started his work at 08:00:00 am. The format of time is HH:MM:SS am|pm.


Sample Input

2
2
20 25
40
1
8



Sample Output

08:00:40 am

08:00:08 am

dp[i]：表示第i个人买完需要的最小时间，转移为dp[i]=min ( dp[i-1]+s[i] , dp[i-2]+d[i] );即前一个人买票和前两个人一起买票的时间花销最小值

#include
using namespace std;
templateinline T read(T&x)
{
    char c;
    while((c=getchar())<=32)if(c==EOF)return 0;
    bool ok=false;
    if(c=='-')ok=true,c=getchar();
    for(x=0; c>32; c=getchar())
        x=x*10+c-'0';
    if(ok)x=-x;
    return 1;
}
template inline T read_(T&x,T&y)
{
    return read(x)&&read(y);
}
template inline T read__(T&x,T&y,T&z)
{
    return read(x)&&read(y)&&read(z);
}
template inline void write(T x)
{
    if(x<0)putchar('-'),x=-x;
    if(x<10)putchar(x+'0');
    else write(x/10),putchar(x%10+'0');
}
templateinline void writeln(T x)
{
    write(x);
    putchar('\n');
}
//-------ZCC IO template------
const int maxn=1e5+1000;
const double inf=999999999;
#define lson (rt<<1),L,M
#define rson (rt<<1|1),M+1,R
#define M ((L+R)>>1)
#define For(i,t,n) for(int i=(t);i<(n);i++)
typedef long long  LL;
typedef double DB;
typedef pair P;
#define bug printf("---\n");
#define mod  100007
int dp[maxn];//第i个人买完票 需要的最小时间
int s[maxn];
int d[maxn];
int main()
{
    int n,E,F;
    read(n);
    while(n--)
    {
        int k;
        read(k);
        for(int i=1;i<=k;i++)
            read(s[i]);

        for(int i=2;i<=k;i++)
            read(d[i]);
        memset(dp,0,sizeof(dp));
        dp[1]=s[1];
        for(int i=2;i<=k;i++)
        {
            dp[i]=min(dp[i-1]+s[i],dp[i-2]+d[i]);
        }

    
        int m=dp[k]/60;
        int h=m/60;
        int ss=dp[k]%60;
        m%=60;
        h+=8;
        if(h<=12)
            printf("%02d:%02d:%02d am\n",h,m,ss);
        else
            printf("%02d:%02d:%02d pm\n",h-12,m,ss);
    }
    return 0;
}