HDOJ 4293 Groups

水DP。。。

Groups

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1133    Accepted Submission(s): 440

Problem Description

　　After the regional contest, all the ACMers are walking alone a very long avenue to the dining hall in groups. Groups can vary in size for kinds of reasons, which means, several players could walk together, forming a group.
　　As the leader of the volunteers, you want to know where each player is. So you call every player on the road, and get the reply like “Well, there are A _i players in front of our group, as well as B _i players are following us.” from the i ^th player.
　　You may assume that only N players walk in their way, and you get N information, one from each player.
　　When you collected all the information, you found that you’re provided with wrong information. You would like to figure out, in the best situation, the number of people who provide correct information. By saying “the best situation” we mean as many people as possible are providing correct information.

 

Input

　　There’re several test cases.
　　In each test case, the first line contains a single integer N (1 <= N <= 500) denoting the number of players along the avenue. The following N lines specify the players. Each of them contains two integers A _i and B _i (0 <= A _i,B _i < N) separated by single spaces.
　　Please process until EOF (End Of File).

 

Output

　　For each test case your program should output a single integer M, the maximum number of players providing correct information.

 

Sample Input

   
   
   
   

    
    
    
    3
2 0
0 2
2 2
3
2 0
0 2
2 2
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    2
2

    
    
    
    
 
     
 
      Hint 
     
 The third player must be making a mistake, since only 3 plays exist. 
    
    
    
    
     
   
   
   
   

 

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

using namespace std;

int n,dp[550],p[550][550];

int main()
{
	while(scanf("%d",&n)!=EOF)	
	{
		int a,b;
		memset(dp,0,sizeof(dp));
		memset(p,0,sizeof(p));
		for(int i=0;i<n;i++) 
		{
			scanf("%d%d",&a,&b);
			int st=a+1,ed=n-b;
			p[st][ed]=min(p[st][ed]+1,ed-st+1);
		}
		for(int i=1;i<=n;i++)
		{
			for(int j=0;j<i;j++)
			{
				dp[i]=max(dp[i],dp[j]+p[j+1][i]);
			}
		}
		printf("%d\n",dp[n]);
	}
	return 0;
}