HDOJ 5125 magic balls DP

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码代码的猿猿的AC之路

最新推荐文章于 2024-10-12 14:52:57 发布

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本文探讨了一道经典的动态规划问题——通过交换魔法球体积来寻找最长递增序列。问题中，每个个体拥有两个不同体积的魔法球，通过消耗能量交换球的体积，求解在给定能量限制下最长递增序列的长度。

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magic balls

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 561 Accepted Submission(s): 168

Problem Description

The town of W has N people. Each person takes two magic balls A and B every day. Each ball has the volume

ai and

bi . People often stand together. The wizard will find the longest increasing subsequence in the ball A. The wizard has M energy. Each point of energy can change the two balls’ volume.(

swap(ai,bi) ).The wizard wants to know how to make the longest increasing subsequence and the energy is not negative in last. In order to simplify the problem, you only need to output how long the longest increasing subsequence is.

Input

The first line contains a single integer

T(1≤T≤20) (the data for

N>100 less than 6 cases), indicating the number of test cases.
Each test case begins with two integer

N(1≤N≤1000) and

M(0≤M≤1000) ,indicating the number of people and the number of the wizard’s energy. Next N lines contains two integer

ai and

bi(1≤ai,bi≤109) ,indicating the balls’ volume.

Output

For each case, output an integer means how long the longest increasing subsequence is.

Sample Input

Sample Output

4
4

Source

BestCoder Round #20

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn=1111;

int n,m;
int a[maxn],b[maxn];
int dp[maxn][maxn][2];

int main()
{
	int T_T;
	scanf("%d",&T_T);
	while(T_T--)
	{
		scanf("%d%d",&n,&m);
		for(int i=1;i<=n;i++)
		{
			scanf("%d%d",a+i,b+i);
		}
		memset(dp,-1,sizeof(dp));
		dp[0][0][0]=0;
		for(int i=1;i<=n;i++)
		{
			dp[i][0][0]=1; dp[i][1][1]=1;
			for(int j=1;j<i;j++)
			{
				int e=min(j,m);
				if(a[i]>a[j])
				{
					for(int k=0;k<=e;k++)
					{
						if(dp[j][k][0]>=0) 
							dp[i][k][0]=max(dp[i][k][0],dp[j][k][0]+1);
					}
				}
				if(a[i]>b[j])
				{
					for(int k=0;k<=e;k++)
					{
						if(dp[j][k][1]>=0) 
							dp[i][k][0]=max(dp[i][k][0],dp[j][k][1]+1);
					}
				}
				if(b[i]>a[j])
				{
					for(int k=0;k<=e;k++)
					{
						if(dp[j][k][0]>=0) 
							dp[i][k+1][1]=max(dp[i][k+1][1],dp[j][k][0]+1);
					}
				}
				if(b[i]>b[j])
				{
					for(int k=0;k<=e;k++)
					{
						if(dp[j][k][1]>=0) 
							dp[i][k+1][1]=max(dp[i][k+1][1],dp[j][k][1]+1);
					}
				}
			}
		}
		int ans=1;
		for(int i=1;i<=n;i++)
		{
			int e=min(i,m);
			for(int j=0;j<=e;j++)
			{
				ans=max(ans,max(dp[i][j][0],dp[i][j][1]));
			}
		}
		printf("%d\n",ans);
	}
	return 0;
}