POJ 2609 Ferry Loading(双塔DP)

Ferry Loading

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1807		Accepted: 509		Special Judge

Description

Before bridges were common, ferries were used to transport cars across rivers. River ferries, unlike their larger cousins, run on a guide line and are powered by the river's current. Two lanes of cars drive onto the ferry from one end, the ferry crosses the river, and the cars exit from the other end of the ferry. 
The cars waiting to board the ferry form a single queue, and the operator directs each car in turn to drive onto the port (left) or starboard (right) lane of the ferry so as to balance the load. Each car in the queue has a different length, which the operator estimates by inspecting the queue. Based on this inspection, the operator decides which side of the ferry each car should board, and boards as many cars as possible from the queue, subject to the length limit of the ferry. Your job is to write a program that will tell the operator which car to load on which side so as to maximize the number of cars loaded. 

Input

The first line of input contains a single integer between 1 and 100: the length of the ferry (in metres). For each car in the queue there is an additional line of input specifying the length of the car (in cm, an integer between 100 and 3000 inclusive). A final line of input contains the integer 0. The cars must be loaded in order, subject to the constraint that the total length of cars on either side does not exceed the length of the ferry. Subject to this constraint as many cars should be loaded as possible, starting with the first car in the queue and loading cars in order until no more can be loaded.

Output

The first line of outuput should give the number of cars that can be loaded onto the ferry. For each car that can be loaded onto the ferry, in the order the cars appear in the input, output a line containing "port" if the car is to be directed to the port side and "starboard" if the car is to be directed to the starboard side. If several arrangements of the cars meet the criteria above, any one will do.

Sample Input

50
2500
3000
1000
1000
1500
700
800
0

Sample Output

6
port
starboard
starboard
starboard
port
port
双塔DP
n个车要么放左边，要么放右边，最大长度不能超过船的长度
这道题目也要记录路径，而且两边的差值有10000，所以必须要用
滚动数组了，要不然内存超限。
dp[i][j] 第i辆车，两边的差距为j
<pre name="code" class="html">#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <stdio.h>
#include <math.h>

using namespace std;
int dp[2][20005];
int d[505];
int e[505];
int a[505];
int ans[505][20005];

int m;
int n;
void dfs(int i,int k)
{
	if(i<=0)
		return;

	if(ans[i][k]<0)
	{
		dfs(i-1,abs(ans[i][k])-2);
		cout<<"port"<<endl;
	}
	else
	{
		dfs(i-1,ans[i][k]);
		cout<<"starboard"<<endl;
	}
}
int main()
{
	scanf("%d",&m);

	m*=100;
	int x;
	scanf("%d",&x);
	int n=0;
	while(x!=0)
	{
		a[++n]=x;
		scanf("%d",&x);
	}
	memset(dp,-1,sizeof(dp));
	memset(ans,-1,sizeof(ans));
	dp[0][0+10000]=0;
	int now=1;
	for(int i=1;i<=n;i++)
		d[i]=10000000;
	for(int i=1;i<=n;i++)
	{
		for(int j=-10000;j<=10000;j++)
		{
			if(dp[now^1][j+10000]>m) continue;
			if(dp[now^1][j+10000]==-1)
				continue;

			if(j<0)
			{
				if(j+10000-a[i]>=0)//不写会re
				{
					dp[now][j+10000-a[i]]=dp[now^1][j+10000]+a[i];
					if(d[i]>dp[now][j+10000-a[i]])
					{
						d[i]=dp[now][j+10000-a[i]];//记录第i辆车形成的最小长度
						e[i]=j+10000-a[i];//第i辆车形成最小长度的差值，
					}

					ans[i][j+10000-a[i]]=(j+10000)*(-1)-2;//记录路径
				}
				dp[now][j+10000+a[i]]=dp[now^1][j+10000]+max(0,j+a[i]);
				if(d[i]>dp[now][j+10000+a[i]])
				{

					d[i]=dp[now][j+10000+a[i]];
					e[i]=j+10000+a[i];
				}


				ans[i][j+10000+a[i]]=j+10000; 
			}
			else
			{
				if(j+10000+a[i]<=20000)
				{
					dp[now][j+10000+a[i]]=dp[now^1][j+10000]+a[i];
					if(d[i]>dp[now][j+10000+a[i]])
					{

						d[i]=dp[now][j+10000+a[i]] ;
						e[i]=j+10000+a[i];
					}
					ans[i][ j+10000+a[i]]=j+10000;
				}

				dp[now][j+10000-a[i]]=dp[now^1][j+10000]+max(0,a[i]-j);
				if(d[i]>dp[now][j+10000-a[i]])
				{
					d[i]=dp[now][j+10000-a[i]];
					e[i]= j+10000-a[i];
				}
				ans[i][j+10000-a[i]]=(j+10000)*(-1)-2; 
			}
		}
		memset(dp[now^1],-1,sizeof(dp[now^1]));
		now^=1;
		if(d[i]>m)
			break;
	}
	int i;
	for(i=n;i>=1;i--)
	{
		if(d[i]<=m)
			break;
	}
	printf("%d
",i);
	dfs(i,e[i]);



	return 0;
}