HDU 1260: Tickets (DP), hdu1260ticketsdp

HDU 1260: Tickets (DP), hdu1260ticketsdp

Tickets Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission (s): 923 Accepted Submission (s): 467


Problem DescriptionJesus, 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 cocould he go back home as early as possible.
A good approach, cing the total time of tickets selling, is let adjacent people buy tickets together. as the restriction of the Ticket Seller Machine, Joe can handle 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. cocould you so kind to tell poor Joe at what time cocould he go back home as early as possible? If so, I guess Joe wocould full of appreciation for your help.
 
InputThere 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 (0 s <= Si <= 25 s) representing the time consumed to buy a ticket for each person;
(K-1) integer numbers (0 s <= Di <= 50 s) representing the time needed for two adjacent people to buy two tickets together.
 
OutputFor every scenario, please tell Joe at what time cocould 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

2220 254018

 
Sample Output

08:00:40 am08:00:08 am

Solution:

A person can separately buy a ticket to spend a certain amount of time, can also be two people to buy a ticket together, also given a time, give K individual separately buy a ticket time and K-1 two adjacent two people to buy a ticket together, ask the minimum time spent in total.

One [I] is the time for each individual to buy a ticket, and two [I + 1] is the time for two people to buy a ticket together.

The state transition equation is dp [I] = min (dp [I-1] + one [I], dp [I-2] + two [I]). Currently, the I-th individual can be purchased separately, but with the previous one.

Pay attention to the final output.

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<stdlib.h>#include<vector>#include<queue>#include<cmath>using namespace std;const int maxn = 2000 + 50;int n;int k;int dp[maxn];int one[maxn];int two[maxn];int main(){    scanf("%d", &n);    while( n-- )    {        memset(dp, 0, sizeof(dp));        memset(one, 0, sizeof(one));        memset(two, 0, sizeof(two));        scanf("%d", &k);        for(int i=1; i<=k; i++)scanf("%d", &one[i]);        for(int i=2; i<=k; i++)scanf("%d", &two[i]);        dp[0] = 0;dp[1] = one[1];        for(int i=2; i<=k; i++)dp[i] = min( dp[i-1]+one[i], dp[i-2]+two[i] );        int h = dp[k] / 3600 + 8;        int m = dp[k] / 60 % 60;        int s = dp[k] % 60;        printf("%02d:%02d:%02d am\n", h, m, s);    }    return 0;}

Hdu1308 http: // acmhdueducn/showproblemphp? Pid = 1, 130

Nclude iostream>
Using namespace std;
# Define MIN (a, B) (a) (B ))? (A) :( B)
# Define MAX (a, B) (a)> (B ))? (A) :( B)
Int point [27];
Char ch [27];
Char d [5001]; // small
Char p [5001];
Int dp [2] [5001];
Int main (){
Int n, I, j, k;
While (scanf ("% d", & n )! = EOF ){
Scanf ("% s", & ch );
Int t;
For (I = 0; in; I ++ ){
Scanf ("% d", & t );
Point [ch [I]-'a'] = t;
}
Scanf ("% s", d + 1 );
Scanf ("% s", p + 1 );
Memset (dp, 0, sizeof (dp ));
Int tmp = 0;
For (I = 1; d [I]; I ++ ){
For (j = 1; p [j]; j ++ ){
If (d [I] = p [j])
Dp [tmp] [j] = dp [! Tmp] [J-1] + point [d [I]-'a'];
Else {
Dp [tmp] [j] = MAX (dp [tmp] [J-1], dp [! Tmp] [j]);
}
}
Tmp =! Tmp;
}
Coutdp [! Tmp] [J-1] endl;
}
Return 0;
}



Let's look at this question. How can I make a mistake? What is HDU's question: http: // acmhdueducn/showproblemphp? Pid = 1, 1565

Your program is not very well-thought-out from the diagonal line. If you don't talk so much about it, You can directly access the data:
3
100 1 1
1 100
200 1 1
You output 303
The correct answer should be 400.