hdu 2881（简单dp）

 题意：n*n的矩阵，里面有m个格子是有任务要去完成的，t,x,y表示要在第t秒到达(x,y)的格子完成任务，问你最多可以完成多少

解题思路：简单dp，将时间排个序后就是LIS

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

const int maxn = 10005;
struct Node
{
	int x,y,t;

	bool operator < (const Node a)
	{
		return t < a.t;
	}
}mission[maxn];
int n,m,dp[maxn];

int main()
{
	while(scanf("%d%d",&n,&m), m || n)
	{
		for(int i = 1; i <= m; i++)
			scanf("%d%d%d",&mission[i].t,&mission[i].x,&mission[i].y);
		sort(mission+1,mission+1+m);
		dp[1] = 1;
		for(int i = 2; i <= m; i++)
		{
			dp[i] = 1;
			for(int j = i - 1; j >= 1; j--)
			{
				if((mission[i].t - mission[j].t) >= (abs(mission[i].x - mission[j].x) + abs(mission[i].y - mission[j].y)))
					dp[i] = max(dp[i],dp[j] + 1);
			}
		}
		int ans = 1;
		for(int i = 1; i <= m; i++)
			ans = max(ans,dp[i]);
		printf("%d\n",ans);
	}
	return 0;
}