Tickets
Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 1935 Accepted Submission (s): 933
Problem Descriptionjesus, what a great movie! Thousands of people is rushing to the cinema. However, the 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, are let adjacent people buy tickets together. As the restriction of the Ticket Seller machine, Joe can sell a single Ticket or both adjacent tickets at a time.
Since you is the great JESUS, you know exactly what much time needed for every person to buy a single ticket or both ticket s for him/her. Could kind to tell poor Joe at what time Could he go home as early as possible? If So, I guess Joe would full of appreciation for your help.
Inputthere is 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 both adjacent people to buy the tickets Togeth Er.
Outputfor every scenario, please tell Joe at-time could he go back home as early as possible. Every day Joe started he work at 08:00:00 am. The format of time is HH:MM:SS am|pm.
Sample Input2220 254018
Sample output08:00:40 am08:00:08 am a long time did not do DP and add their own DP is very slag, the afternoon game to see people a one to make, their own can only helplessly look, alas!!! IQ AH!! Test instructions: A group of people to buy tickets, first enter each individual ticket to spend the time, in giving two people 22 combined with the time spent buying tickets to find the shortest time: the need to introduce the state transfer equation, set the array a[] is a single person to buy a ticket to spend time, array b[] is two people spend time to buy a ticket, Dp[i] Indicates the time taken by the former I individual to buy a ticket, then the state transfer equation is: Dp[i]=min (Dp[i-1]+a[i],dp[i-2]+b[i]);
#include <stdio.h> #include <string.h> #define MAX 2100#define min (x, y) (x<y?x:y) int a[max],b[max],dp[ Max];int Main () {int t,i,j,n;int h,m,s;int sum,tot;scanf ("%d", &t), while (t--) {memset (dp,0,sizeof (DP)); SCANF ("%d" , &n), for (i=1;i<=n;i++) scanf ("%d", &a[i]), and for (i=2;i<=n;i++) scanf ("%d", &b[i]);DP [1]=a [1]; for (i=2;i<=n;i++) dp[i]=min (Dp[i-1]+a[i],dp[i-2]+b[i]); printf ("%d\n", Dp[n]); Sum=dp[n]; h=0;s=0;m=0; s=sum%60; M= (sum-s)/60; if (m>=60) { h=h+m/60; m=m%60; } H=8+h; if (h<=12) printf ("%02d:%02d:%02d am\n", h,m,s); else { h-=12; printf ("%02d:%02d:%02d pm\n", h,m,s); }
HDOJ 1260 Tickets "DP"
