nyoj 716 River Crossing 【多决策 DP】

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

理解题意后超简单的：一遍就过了。   注意： 题中所给的Mi是指带 i 只羊所需时间，不是带第 i 只羊所需时间。

#include<stdio.h>
#include<string.h>
#define INF 0x3f3f3f
#define max 1000+10
int M[max];
int time[max];//time[i]数组存储带i只羊花费时间 
int dp[max];//dp[i]存储带i只羊所需最优时间 
int min(int x,int y)
{
	return x<y?x:y;
}
int main()
{
	int t,n,m;
	int i,j,sum;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&m);
		memset(time,0,sizeof(time));
		for(i=1;i<=n;i++)
		{
			sum=0;
			scanf("%d",&M[i]);
			for(j=1;j<=i;j++)
			{
				sum+=M[j];
			}
			time[i]=sum;
		}
		dp[1]=time[1]+m;
		dp[0]=0;
		for(i=2;i<=n;i++)
		{
			dp[i]=INF;
			for(j=0;j<=i;j++)
			{
				if(j>0)//一开始就全部带走  不需要返回时间 
				dp[i]=min(dp[i],dp[j]+m+time[i-j]+m);
				else
				dp[i]=min(dp[i],dp[j]+time[i-j]+m);
			}
		}
		printf("%d\n",dp[n]);
	} 
	return 0;
}