题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6447

YJJ's Salesman

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2062 Accepted Submission(s): 776

Problem Description

YJJ is a salesman who has traveled through western country. YJJ is always on journey. Either is he at the destination, or on the way to destination.
One day, he is going to travel from city A to southeastern city B. Let us assume that A is (0,0) on the rectangle map and B (109,109) . YJJ is so busy so he never turn back or go twice the same way, he will only move to east, south or southeast, which means, if YJJ is at (x,y) now (0≤x≤109,0≤y≤109) , he will only forward to (x+1,y) , (x,y+1) or (x+1,y+1) .
On the rectangle map from (0,0) to (1e9,1e9) , there are several villages scattering on the map. Villagers will do business deals with salesmen from northwestern, but not northern or western. In mathematical language, this means when there is a village k on (xk,yk) (1≤xk≤109,1≤yk≤109) , only the one who was from (xk−1,yk−1) to (xk,yk) will be able to earn vk dollars.(YJJ may get different number of dollars from different village.)
YJJ has no time to plan the path, can you help him to find maximum of dollars YJJ can get.

Input

The first line of the input contains an integer T (1≤T≤10) ,which is the number of test cases.

In each case, the first line of the input contains an integer N (1≤N≤105) .The following N lines, the k -th line contains 3 integers, xk,yk,vk (0≤vk≤103) , which indicate that there is a village on (xk,yk) and he can get vk dollars in that village.
The positions of each village is distinct.

Output

The maximum of dollars YJJ can get.

Sample Input

1 3 1 1 1 1 2 2 3 3 1

Sample Output

题目大意：从(0,0)走到(1e9,1e9)，假设此刻在(xk,yk)，那么只能走到(xk+1,yk)，(xk,yk+1)，(xk+1,yk+1)，且走到(xk+1,yk+1)时能获得(xk,yk)的值vk，求到达终点时的最大值

思路：这题本身思路不难，简单的dp问题，状态转移方程为dp[x][y] = max(dp[x-1][y],dp[x][y-1],dp[x-1][y-1]+v[x-1][y-1])。但关键是，这题数据量很大，直接建图dp连内存都不够。所幸题目vk>0的点最多只有1e5，所以需要用到离散化。离散化后，每给点都需要向前查找最大值，所以O(n^2)，利用树状数组可以减到O(nlogn)。

附上AC代码：

#include<bits/stdc++.h>
#define ll long long
using namespace std;
struct node
{
	int x,y,value;
}mp[100005];
int sum[100005],a[100005],t,n;
bool cmp(node a ,node b)
{
	if(a.x != b.x)
		return a.x < b.x;
	return a.y > b.y;
}
int lowbit(int x)
{
	return x&(-x);
}
void update(int p,int val)
{
	for(int i = p;i <= n;i+=lowbit(i))
		sum[i] = max(sum[i],val);
}
int query(int p)
{
	int ans = 0;
	for(int i = p;i >= 1;i-=lowbit(i))
		ans = max(sum[i],ans);
	return ans;	
}
int main()
{
	while(scanf("%d",&t) !=EOF)
	{
		while(t--)
		{
			memset(sum,0,sizeof(sum));
			scanf("%d",&n);
			for(int i = 1;i <= n;i++)
			{
				scanf("%d%d%d",&mp[i].x,&mp[i].y,&mp[i].value);
				a[i] = mp[i].y;//a[i]将y离散化
			}
			sort(a+1,a+1+n);
			int all = unique(a+1,a+1+n)-a-1;//去重
			sort(mp+1,mp+1+n,cmp);
			for(int i = 1;i <= n;i++)
			{ 
				int Y = lower_bound(a+1,a+1+all,mp[i].y)-a;//查找当前y的大小对应的位置
				int val = query(Y-1)+mp[i].value;
				update(Y,val);
			}
			cout << query(n) << endl;
		}
	}
	//cout << "AC" << endl;
	return 0;
}

2018网络赛CCPC——YJJ's Salesman（离散化+dp+树状数组/线段树）

YJJ's Salesman

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