HDU
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Tickets
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.
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.
2220 254018
08:00:40 am08:00:08 am
题意:
买票。可以一个人买,也可以两个人买,问怎样买可以花费最少
(选择性问题也能用动态规划的思想解决)
代码:
#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>using namespace std;#define INF 0x3f3f3f3fvoid print(int s){int H=8,M=0,S=0;S+=s;if(S>=60){M=S/60;S=S-M*60;}if(M>=60){int t=M/60;H += t;M=M-t*60;}printf("%02d:%02d:%02d %s\n",H,M,S,H>=12?"pm":"am");}int main(){int n,tx;int s[2005],d[2005];int dp[2005];scanf("%d",&tx);while(tx--){scanf("%d",&n);memset(dp,0,sizeof(dp));for(int i=1;i<=n;i++)scanf("%d",&s[i]);for(int i=2;i<=n;i++)scanf("%d",&d[i]);dp[1]=s[1];dp[2]=min(s[1]+s[2],d[2]);for(int i=3;i<=n;i++){dp[i]=min(dp[i-1]+s[i],dp[i-2]+d[i]);}print(dp[n]);}return 0;}