hdu 6201 深度优先搜索_hdu 深度优先搜索-优快云博客

本文介绍了一道关于商人如何在不同城市间买卖图书以获得最大利润的问题。商人需考虑各城市图书价格及城市间的旅行成本。

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

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

真的，要不是当时评测机被那帮交脚本做题的家伙搞炸了，这题肯定能在比赛时做出来。

注意：这题时间限制是2000ms！！！

说一下思路：

这题的意思就是：一个商人只能在一个城市内买书或买书，然后把所买的书卖到下一个城市，而且买和卖都只有一次，期间从一个城市到另一个城市的路径长度要当成路费，问这个买卖的最大利润是多少。

那么这道题就是要你求一个点的点权，与另一个点的点权与两点之间的权值（中间可能隔着几个点）的和，的差的最大值。

即是：

maxx = max(maxx, points[maps[u][i].v] - points[st] - maps[u][i].w - dis);
maxx = max(maxx, points[st] - points[maps[u][i].v] - maps[u][i].w - dis);

这里，maxx是指那个最大值，st 指起点，maps[u][i].v 指终点，u 指终点的前一个点，i 指 u 点连接点的个数，maps[u][i].w 指 u 到 v 的边的权值，dis 指起点到 u 点的所有边的权值和，points 数组就是记录所有点的权值

maps在这里是vector，其声明为：

struct node
{
LL v, w;
};
vector<node>maps[maxn];

要是vector不会的话百度一下吧

最后附上代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long LL;
///o(っ。Д。)っ AC万岁!!!!!!!!!!!!!!
const int maxn = 100090;
const int inf = -999999999;
struct node
{
int v, w;
};
vector<node>maps[maxn];
int points[maxn] = {}, maxx;
void dfs(int st, int u, int uv, int dis)
{
for(int i = 0; i < maps[u].size(); i++)
{
if(maps[u][i].v == st || maps[u][i].v == uv) continue;
maxx = max(maxx, points[maps[u][i].v] - points[st] - maps[u][i].w - dis);
maxx = max(maxx, points[st] - points[maps[u][i].v] - maps[u][i].w - dis);
dfs(st, maps[u][i].v, u, dis + maps[u][i].w);
}
}
int main()
{
int _;
scanf("%d", &_);
while(_--)
{
int n;
int u, v, w;
scanf("%d", &n);
struct node nn;
for(int i = 1; i <= n; i++)
{
scanf("%d", &points[i]);
}
for(int i = 1; i < n; i++)
{
scanf("%d %d %d", &u, &v, &w);
if(u > v)
{
swap(u, v);
}
nn.v = v;
nn.w = w;
maps[u].push_back(nn);
}
maxx = inf;
for(int i = 1; i <= n; i++)
{
dfs(i, i, i, 0);
}
for(int i = 1; i <= n; i++)
{
maps[i].clear();
}
printf("%d\n", maxx);
}
return 0;
}
/*
4
30 40 10 40
1 3 20
1 2 40
1 4 10
*/