River Crossing
来源:互联网 发布:c语言考研题库 编辑:程序博客网 时间:2024/05/25 20:00
River Crossing
时间限制:1000 ms | 内存限制:65535 KB
难度:4
- 描述
Afandi is herding N sheep across the expanses of grassland when he finds himself blocked by a river. A single raft is available for transportation.
Afandi knows that he must ride on the raft for all crossings, but adding sheep to the raft makes it traverse the river more slowly.
When Afandi is on the raft alone, it can cross the river in M minutes When the i sheep are added, it takes Mi minutes longer to cross the river than with i-1 sheep (i.e., total M+M1 minutes with one sheep, M+M1+M2 with two, etc.).
Determine the minimum time it takes for Afandi to get all of the sheep across the river (including time returning to get more sheep).
- 输入
- On the first line of the input is a single positive integer k, telling the number of test cases to follow. 1 ≤ k ≤ 5 Each case contains:
* Line 1: one space-separated integers: N and M (1 ≤ N ≤ 1000 , 1≤ M ≤ 500).
* Lines 2..N+1: Line i+1 contains a single integer: Mi (1 ≤ Mi ≤ 1000) - 输出
- For each test case, output a line with the minimum time it takes for Afandi to get all of the sheep across the river.
- 样例输入
2
2 10
3
5
5 10
3
4
6
100
1
- 样例输出
18
50
/* 分类:DP NYOJ river crossing 题意:有1个人和N只羊要过河。一个人单独过河花费的时间是M,每次带一只羊过河花费时间M+M1,带两只羊过河花费时间M+M1+M2……给出N、M和Mi,问N只羊全部过河最少花费的时间是多少。分析:用一个前缀和数组time,time[i]表示单独运送i只羊所花费的时间。dp[i]表示一个人和i只羊过河所花费的最短时间,则开始时dp[i] = time[i] + M,以后更新时,dp[i] = min(dp[i],dp[i-j] + m + dp[j]),j从1循环到i-1,即把i只羊分成两个阶段来运,只需求出这两个阶段的和,然后加上人从对岸回来所用的时间,与dp[i]进行比较,取最小值。 */#include<stdio.h>#include<algorithm>#include<string.h>using namespace std;int main(){int k,n,m;int a[1004];int dp[1004];scanf("%d",&k);while(k--){scanf("%d%d",&n,&m);memset(dp,0,sizeof(dp));for(int i=1;i<=n;i++){scanf("%d",&a[i]);dp[i]=a[i]+m;}for(int i=2;i<=n;i++)for(int j=1;j<i;j++)dp[i]+=a[j];/*for(int i=1;i<=n;i++)printf("%d\n",dp[i]);*/for(int i=2;i<=n;i++){//j要取i/2,取一半就可以了,不是j*j<i** ║** for(int j=1;j<=i/2;j++){dp[i]=min(dp[i],dp[j]+dp[i-j]+m);}}printf("%d\n",dp[n]);}}
0 0
- Crossing River
- Crossing River
- Crossing River
- Crossing River
- Crossing River
- River Crossing
- Crossing River
- PKU 1700 Crossing River
- River Crossing解题报告
- pku 1700 Crossing River
- POJ1700 Crossing River DP
- Crossing River POJ1700
- poj 1718 River Crossing
- poj 1700 Crossing River
- Crossing River (P1700)
- poj1700 Crossing River 贪心
- Poj 1700 Crossing River
- POJ 1700 Crossing River
- code[vs] 1474十进制-m进制
- Linux sed之删除文件第一行
- 【SSH进阶之路】Struts基本原理 + 实现简单登录(二)
- 如何写死 u-boot 中的 bootargs
- CSDN-栏目美化
- River Crossing
- 【SSH进阶之路】一步步重构MVC实现Struts框架——从一个简单MVC开始(三)
- bpm系统渠道线索页面静态转动态
- Android 5.0 6.0 以及7.0新特性 MD风格 敏感权限 文件访问
- nginx的完整配置示例
- selenium使用遇到的问题
- 带返回值的多线程实现
- springMVC的全流程使用和分析
- 使用bootstrapvalidator的remote验证经验