Narrow Art Gallery

Time Limit: 4000ms, Special Time Limit:10000ms, Memory Limit:65536KB

Total submit users: 11, Accepted users: 9

Problem 13240 : No special judgement

Problem description

A long art gallery has 2N rooms. The gallery is laid out as N rows of 2 rooms side-by-side. Doors connect all adjacent rooms (north-south and east-west, but not diagonally). The curator has been told that she must close off k of the rooms because of staffing
cuts. Visitors must be able to enter using at least one of the two rooms at one end of the gallery, proceed through the gallery, and exit from at least one of the two rooms at the other end. Therefore, the curator must not close off any two rooms that would
block passage through the gallery. That is, the curator may not block off two rooms in the same row or two rooms in adjacent rows that touch diagonally. Furthermore, she has determined how much value each room has to the general public, and now she wants to
close off those k rooms that leave the most value available to the public, without blocking passage through the gallery.

Input

Input will consist of multiple problem instances (galleries). Each problem instance will begin with a line containing two integers N and k, where 3 ≤ N ≤ 200 gives the number of rows, and 0 ≤ k ≤ N gives the number of rooms that must be closed off. This
is followed by N rows of two integers, giving the values of the two rooms in that row. Each room’s value v satisfies 0 ≤ v ≤ 100. A line containing 0 0 will follow the last gallery.

Output

For each gallery, output the amount of value that the general public may optimally receive, one line per gallery.

Sample Input

6 4
3 1
2 1
1 2
1 3
3 3
0 0

4 3
3 4
1 1
1 1
5 6

10 5
7 8
4 9
3 7
5 9
7 2
10 3
0 10
3 2
6 3
7 9

0 0

Sample Output

17
17
102

Problem Source

ACM-ICPC North America Qualifier 2014

我还是非常弱的……

dp[i][j][k]:前i行在状态j时，关掉k个房间后的最大值

当然有些k在某些状态时是訪问不到的，所以此时赋上一个非常小的值。这样就不影响结果

j是有3个状态。0为i行两个都取不到。1为关掉i行左边那个房间，2为关掉i行右边那个房间

#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#include<bitset>
#include<climits>
#include<list>
#include<iomanip>
#include<stack>
#include<set>
using namespace std;
int dp[210][3][210],k,n,g[210][2];
int work()
{
	for(int i=0;i<3;i++)
		fill(dp[0][i],dp[0][i]+k+1,INT_MIN);
	dp[0][0][0]=g[0][0]+g[0][1];
	dp[0][1][1]=g[0][1];
	dp[0][2][1]=g[0][0];
	for(int i=1;i<n;i++)
		for(int j=0;j<=k;j++)
		{
			dp[i][0][j]=max(dp[i-1][0][j],max(dp[i-1][1][j],dp[i-1][2][j]))+g[i][0]+g[i][1];
			if(j==0)
				dp[i][1][j]=dp[i][2][j]=INT_MIN;
			else
			{
				dp[i][1][j]=max(dp[i-1][0][j-1],dp[i-1][1][j-1])+g[i][1];
				dp[i][2][j]=max(dp[i-1][0][j-1],dp[i-1][2][j-1])+g[i][0];
			}
		}
	return max(dp[n-1][0][k],max(dp[n-1][1][k],dp[n-1][2][k]));
}
int main()
{
	while(cin>>n>>k)
	{
		if(n==0&&k==0)
			break;
		for(int i=0;i<n;i++)
			for(int j=0;j<2;j++)
				cin>>g[i][j];
		cout<<work()<<endl;
	}
}