HDU 1260 Tickets(DP)
2017-07-23 10:58
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Tickets
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4513 Accepted Submission(s): 2320
Problem 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
Source
浙江工业大学第四届大学生程序设计竞赛
Recommend
JGShining
题意:有N个人排队买票,可以一个人买,可以两相邻的人一起买,给出每个人买票的时间和每两个相邻的人买票的时间,使得买票的总时间最短。
做法:dp
dp[i]表示前i个人买票的最短时间且只有i个人,则对于第i个人来说,可以单独买,可以和他前面的人一起买
dp[i]=min(dp[i-2]+aj[i-1],dp[i-1]+s[i])
和那题删一个数字取它相邻的数字很相像,这是按照现阶段决策确定的dp
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4513 Accepted Submission(s): 2320
Problem 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
Source
浙江工业大学第四届大学生程序设计竞赛
Recommend
JGShining
题意:有N个人排队买票,可以一个人买,可以两相邻的人一起买,给出每个人买票的时间和每两个相邻的人买票的时间,使得买票的总时间最短。
做法:dp
dp[i]表示前i个人买票的最短时间且只有i个人,则对于第i个人来说,可以单独买,可以和他前面的人一起买
dp[i]=min(dp[i-2]+aj[i-1],dp[i-1]+s[i])
和那题删一个数字取它相邻的数字很相像,这是按照现阶段决策确定的dp
#include <iostream> #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #define maxn 10010 using namespace std; const int inf=0x3f3f3f3f3f; int dp[maxn]; int aj[maxn]; int s[maxn]; int main() { int t; scanf("%d",&t); int n; while(t--) { scanf("%d",&n); for(int i=1;i<=n;i++) { scanf("%d",&s[i]); } memset(dp,inf,sizeof(dp)); dp[0]=0; for(int i=1;i<=n-1;i++) { scanf("%d",&aj[i]); } dp[1]=s[1]; for(int i=2;i<=n;i++) { dp[i]=min(dp[i-2]+aj[i-1],dp[i-1]+s[i]); } int ans=dp ; int h=ans/3600; int m=(ans/60)%60; int s=ans%60; h+=8; if(h<10)printf("0%d:",h);else printf("%d:",h); if(m<10)printf("0%d:",m);else printf("%d:",m); if(s<10)printf("0%d",s);else printf("%d",s); if(h>12)printf(" pm\n");else printf(" am\n"); } return 0; }
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