hdu 6447 YJJ's Salesman（線段樹，離散化優化dp） (2018CCPC 網絡選拔賽 )

YJJ's Salesman

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1524    Accepted Submission(s): 546

 

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 (109,109) , 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

3

 

題意：

yjj從A（0,0）走到B（1e9,1e9），中間有n個村莊，從村莊的西北方向走向村莊可以賺取v美元（如(x-1,y-1)-->(x,y)）

求能獲得的最高金額；

 

思路，將村莊按x升序y降序排序，對y座標離散化，然後對於第k個村莊（xk,yk）,用線段樹先記錄k-1個村莊是在各個y上能獲得的最大金額（因爲按x升序，只需考慮y），查詢y在[1，yk-1]間的最大值maxz，加上vk，更新線段樹。

最後將答案輸出；

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include<algorithm>
#include<iostream>
#include<set>
#include<map>
#include<queue>
#include<stack>
using namespace std;
const int N = 100005;
typedef long long ll;
const ll mod = 1000000007;
struct my
{
	int x,y,c;
	bool operator<(const my a)const
	{
		if(a.x==x)return a.y<y;
		return a.x>x;
	}
}a[N];
int ly[N],lc[N];
int ax[N],ay[N];
int q[N<<2];
void updata(int p,int l,int r,int x,int y)
{
	if(l==r)
	{
		q[p]=max(q[p],y);
		return;
	}
	int mid=(l+r)>>1;
	if(mid>=x)updata(p<<1,l,mid,x,y);
	else updata(p<<1|1,mid+1,r,x,y);
	q[p]=max(q[p<<1],q[p<<1|1]);
}
int query(int p,int l,int r,int x,int y)
{
	if(l==x&&r==y)return q[p];
	int mid=(l+r)>>1;
	if(mid>=y)return query(p<<1,l,mid,x,y);
	else if(mid<x)return query(p<<1|1,mid+1,r,x,y);
	else return max(query(p<<1,l,mid,x,mid),query(p<<1|1,mid+1,r,mid+1,y));
}
void init(int p,int l,int r)
{
	q[p]=0;
	if(l==r)return ;
	int mid=(l+r)>>1;
	init(p<<1,l,mid);
	init(p<<1|1,mid+1,r);
}
int main()
{
	int t,n;
	scanf("%d",&t);
	while(t--)
	{
	
		int x,y,c;
		int xlen=0,ylen=0;
		scanf("%d",&n);
		for(int i=0;i<n;i++)
		{
			scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].c);
			//	ax[xlen++]=a[len].x;
				ay[ylen++]=a[i].y;
		}
	    sort(a,a+n);
		//sort(ax,ax+xlen);
	//	xlen=unique(ax,ax+xlen)-ax;
		sort(ay,ay+ylen);
		ylen=unique(ay,ay+ylen)-ay;
		for(int i=0;i<n;i++)
		{
			//a[i].x=lower_bound(ax,ax+xlen,a[i].x)-ax+1;
			a[i].y=lower_bound(ay,ay+ylen,a[i].y)-ay+1;
		}
		int u=0;
		int lz=0;
		while(u<n)
		{
			int maxz=0;
			if(a[u].y>1) maxz = query(1,1,ylen,1,a[u].y-1);
			updata(1,1,ylen,a[u].y,maxz+a[u].c);
			u++;
		}

		printf("%d\n",q[1]);
		init(1,1,ylen);
	}
	return 0;
}