HDU6201-transaction transaction transaction

transaction transaction transaction

                                                                     Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others)
                                                                                             Total Submission(s): 1514    Accepted Submission(s): 731

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 ai yuan in it city. Kelukin will take taxi, whose price is 1yuan 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 ith number means the prices in ith 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 zkm (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 2 301 3 23 4 10

 

Sample Output

8

 

Source

2017 ACM/ICPC Asia Regional Shenyang Online

 

题意：给出一棵生成树，每个点有一个权值，代表商品的售价，树上每一条边上也有一个权值，代表从这条边经过所需要的花费。在树上选择两个点，一个作为买入商品的点，一个作为卖出商品的点，当然需要考虑从买入点到卖出点经过边的花费。使得收益最大。允许买入点和卖出点重合，即收益最小值为0

解题思路：

方法一：可以建一个超级源点，从这个超级源点向其他点连边，边的权值为点的权值，求出到各个点的最短路即可

方法二：树形dp，dp[i][0]表示从根节点走到i及其子树并中任一点买入一本书后这个人身上钱的最大值。dp[i][1]表示从根节点走到i及其子树并中任一点卖出一本书后这个人身上钱的最大值。那么对这棵树进行一次树形DP即可，dfs后对每个节点更新收益最大值，单点的计算方法为ans=max(ans,dp[i][0]+dp[i][1])

方法一：最短路

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <cmath>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst LL INF = 0x3f3f3f3f3f3f3f3f;LL dis[100009],l[400009],w,a[100009];int s[100009],nt[400009],e[400009],vis[100009];int n,u,v;struct node{    int id;    LL dis;    bool operator<(const node &a)const    {        return dis>a.dis;    }} pre,nt1;void Dijkstra(){    memset(dis,INF,sizeof dis);    memset(vis,0,sizeof vis);    pre.id=pre.dis=0;    dis[0]=0;    priority_queue<node>q;    q.push(pre);    while(!q.empty())    {        pre=q.top();        q.pop();        vis[pre.id]=1;        for(int i=s[pre.id]; ~i; i=nt[i])        {            if(vis[e[i]]) continue;            if(dis[e[i]]>pre.dis+l[i])            {                dis[e[i]]=pre.dis+l[i];                nt1.id=e[i],nt1.dis=dis[e[i]];                q.push(nt1);            }        }    }    LL ans=0;    for(int i=1; i<=n; i++)    {        LL temp=max(a[i]-dis[i],1LL*0);        ans=max(ans,temp);    }    printf("%lld\n",ans);}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        int cnt=0;        memset(s,-1,sizeof s);        for(int i=1; i<=n; i++)        {            scanf("%lld",&a[i]);            nt[cnt]=s[i],s[i]=cnt,e[cnt]=0,l[cnt++]=a[i];            nt[cnt]=s[0],s[0]=cnt,e[cnt]=i,l[cnt++]=a[i];        }        for(int i=1; i<n; i++)        {            scanf("%d%d%lld",&u,&v,&w);            nt[cnt]=s[u],s[u]=cnt,e[cnt]=v,l[cnt++]=w;            nt[cnt]=s[v],s[v]=cnt,e[cnt]=u,l[cnt++]=w;        }        Dijkstra();    }    return 0;}

方法二：树形dp

#include <iostream>  #include <cstdio>  #include <cstring>  #include <string>  #include <algorithm>  #include <cmath>  #include <map>  #include <set>  #include <stack>  #include <queue>  #include <vector>  #include <bitset>  #include <functional>  using namespace std;#define LL long long  const int INF = 0x3f3f3f3f;int n;LL a[100009], dp[100009][2], v[200009],ans;int s[100009], nt[200009], e[200009];void dfs(int k, int fa){dp[k][0] = -a[k], dp[k][1] = a[k];for (int i = s[k]; ~i; i = nt[i]){if (e[i] == fa) continue;dfs(e[i], k);dp[k][0] = max(dp[k][0], dp[e[i]][0] - v[i]);dp[k][1] = max(dp[k][1], dp[e[i]][1] - v[i]);}ans = max(dp[k][1] + dp[k][0], ans);}int main(){//freopen("input.txt", "r", stdin);  //freopen("output.txt", "w", stdout);int t;scanf("%d", &t);while (t--){scanf("%d", &n);int cnt = 0;memset(s, -1, sizeof s);for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);int uu, vv;LL ww;for (int i = 1; i < n; i++){scanf("%d%d%lld", &uu, &vv, &ww);nt[cnt] = s[uu], s[uu] = cnt, e[cnt] = vv, v[cnt++] = ww;nt[cnt] = s[vv], s[vv] = cnt, e[cnt] = uu, v[cnt++] = ww;}ans = 0;dfs(1,1);printf("%lld\n", ans);}return 0;}