題干：

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)(0,0)?on the rectangle map and B?(109,109)(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)(x,y)?now?(0≤x≤109,0≤y≤109)(0≤x≤109,0≤y≤109), he will only forward to?(x+1,y)(x+1,y),?(x,y+1)(x,y+1)?or?(x+1,y+1)(x+1,y+1).?
On the rectangle map from?(0,0)(0,0)?to?(109,109)(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?kk?on?(xk,yk)(xk,yk)?(1≤xk≤109,1≤yk≤109)(1≤xk≤109,1≤yk≤109), only the one who was from?(xk?1,yk?1)(xk?1,yk?1)?to?(xk,yk)(xk,yk)?will be able to earn?vkvk?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?TT?(1≤T≤10)(1≤T≤10),which is the number of test cases.?

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

題目大意：

商人從（0,0）走到（1e9，1e9）途中有若干村莊，商人可以向（x，y+1）（x+1, y）（x+1, y+1）三個方向前進，當向（x+1,y+1）方向前進時可以獲得村莊（x+1, y+1）的利潤，問商人途中最多可以掙多少錢

解題報告：

商人到達點（x,y）的最大利潤一定是從（0,0）到（x-1，y-1）這個矩形轉移而來的，但是如果單純的用二維來dp會超空間，可以考慮dp的順序而降低dp維度，dp順序從上到下，從右到左，這樣dp[i] = max(dp[k]) + val[i][j] 1 < K < i這樣就降低了一維，用樹狀數組來處理前綴最大值。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define F first #define S second #define ll long long #define pb push_back #define pm make_pair using namespace std; typedef pair<int,int> PII; const int MAX = 2e5 + 5; struct Point {int x,y,v;bool operator <(const Point & b) const {if(x != b.x) return x < b.x;else return y > b.y;} } p[MAX]; int b[MAX],c[MAX],ans,n; int get(int x) {return lower_bound(b+1,b+n+1,x) - b; } int query(int x) {int res = 0;while(x > 0) {res = max(res,c[x]);x -= x&-x;}return res; } void update(int x,int val) {while(x < MAX) {c[x] = max(c[x],val);x += x&-x;} } int main() {int t;cin>>t;while(t--) {cin>>n;ans=0;for(int i = 1; i<=n; i++) scanf("%d%d%d",&p[i].x,&p[i].y,&p[i].v),b[i] = p[i].y,c[i] = 0;sort(b+1,b+n+1);sort(p+1,p+n+1);for(int i = 1; i<=n; i++) {int pos = get(p[i].y);int val = query(pos-1) + p[i].v;ans = max(ans,val);update(pos,val);}printf("%d\n",ans);}return 0 ; }

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