HDU 6201 transaction transaction transaction dp

最新推荐文章于 2019-07-16 21:07:00 发布

LZK1997

最新推荐文章于 2019-07-16 21:07:00 发布

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本文描述了一个商人如何在不同价格的城市间买卖书籍以获取最大利润的问题。通过一次旅行选择两个城市进行交易，考虑路费成本，利用动态规划算法求解最优路径。

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transaction transaction transaction

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 132768/132768 K (Java/Others)
Total Submission(s): 1269 Accepted Submission(s): 609

Problem Description

Kelukin is a businessman. Every day, he travels around cities to do some business. On August 17th, in memory of a great man, citizens will read a book named "the Man Who Changed China". Of course, Kelukin wouldn't miss this chance to make money, but he doesn't have this book. So he has to choose two city to buy and sell.
As we know, the price of this book was different in each city. It is

yuan in

t city. Kelukin will take taxi, whose price is

yuan per km and this fare cannot be ignored.
There are

n−1 roads connecting

n cities. Kelukin can choose any city to start his travel. He want to know the maximum money he can get.

Input

The first line contains an integer

T (

1≤T≤10 ) , the number of test cases.
For each test case:
first line contains an integer

n (

2≤n≤100000 ) means the number of cities;
second line contains

n numbers, the

th number means the prices in

th city;

(1≤Price≤10000)
then follows

n−1 lines, each contains three numbers

x ,

y and

z which means there exists a road between

x and

y , the distance is

(1≤z≤1000) .

Output

For each test case, output a single number in a line: the maximum money he can get.

Sample Input

  
  
   
   1  
4  
10 40 15 30  
1 2 30
1 3 2
3 4 10

Sample Output

Source

2017 ACM/ICPC Asia Regional Shenyang Online

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题意很简单，就是从这个城市进货，去另外一个城市卖，路费算进去书的成本

dp:

dp[nn[i].end]=max(dp[nn[i].end],dp[nn[i].star]-a[nn[i].star]-nn[i].road+a[nn[i].end]);

ac代码：

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <string>

using namespace std;
int t;
int n;
int a[100005];
int dp[100005];
struct node{
	int star,end;
	int road;
}nn[200005];

int main(){
	
	scanf("%d",&t);
	while(t--){
		scanf("%d",&n);
		for(int i=1;i<=n;i++){
			scanf("%d",&a[i]);
		}
		int x,y,z;
		int cnt=0;
		for(int i=1;i<=n-1;i++){
			scanf("%d%d%d",&x,&y,&z);
			nn[cnt].star=x;
			nn[cnt].end=y;
			nn[cnt++].road=z;
			
			nn[cnt].star=y;
			nn[cnt].end=x;
			nn[cnt++].road=z;
			
		}
		memset(dp,0,sizeof(dp));
		int mmax=0;
		for(int i=0;i<cnt;i++){
			dp[nn[i].end]=max(dp[nn[i].end],dp[nn[i].star]-a[nn[i].star]-nn[i].road+a[nn[i].end]);
			mmax=max(mmax,dp[nn[i].end]);
		}
		printf("%d\n",mmax);
	}
	
	
	return 0;
}