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hdu 5834 Magic boy Bi Luo with his excited tree 树形dp+转移

2024-08-13 08:58:25   225人阅读

Magic boy Bi Luo with his excited tree

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1058    Accepted Submission(s): 308


Problem Description
Bi Luo is a magic boy, he also has a migic tree, the tree has N nodes , in each node , there is a treasure, it‘s value is V[i], and for each edge, there is a cost C[i], which means every time you pass the edge i , you need to pay C[i].

You may attention that every V[i] can be taken only once, but for some C[i] , you may cost severial times.

Now, Bi Luo define ans[i] as the most value can Bi Luo gets if Bi Luo starts at node i.

Bi Luo is also an excited boy, now he wants to know every ans[i], can you help him?
 

 

Input
First line is a positive integer T(T≤104) , represents there are T test cases.

Four each test：

The first line contain an integer N(N≤105).

The next line contains N integers V[i], which means the treasure’s value of node i(1≤V[i]≤104).

For the next N−1 lines, each contains three integers u,v,c , which means node u and node v are connected by an edge, it‘s cost is c(1≤c≤104).

You can assume that the sum of N will not exceed 106.
 

 

Output
For the i-th test case , first output Case #i: in a single line , then output N lines , for the i-th line , output ans[i] in a single line.
 

 

Sample Input
154 1 7 7 7 1 2 61 3 12 4 83 5 2
 

 

Sample Output
Case #1:151014915
 

 

Author
UESTC
 

 

Source
2016中国大学生程序设计竞赛 - 网络选拔赛
题意：给你一棵树,每个结点有个宝藏价值w,每次只能拿一次宝藏,每次经过路径需要花费val值,路径可以来回经过,同时可以多次花费val值,求从i点出发,能拿到的最大权值ans【i】
#include <cstdio>#include <iostream>#include <algorithm>#include <cstring>#include <iostream>#include <cmath>#include <map>#include <stack>#include <queue>#include <vector>#include <bitset>#include <set>#define MM(a,b) memset(a,b,sizeof(a));#define inf 0x3f3f3f3fusing namespace std;typedef long long ll;#define CT continue#define SC scanfconst int N=1e5+10;int val[N],dp[N][6],ans[N];struct edge{   int v,c;};vector<edge> G[N];void dfs1(int u,int f){    dp[u][1]=dp[u][0]=val[u];    for(int i=0;i<G[u].size();i++){        int v=G[u][i].v,c=G[u][i].c;        if(v==f) CT;        dfs1(v,u);        int cur=dp[u][0]-c+dp[v][1],            ano1=dp[u][1]+max(dp[v][0]-2*c,0),            ano2=dp[u][2]+max(dp[v][0]-2*c,0);        if(cur>ano1){            dp[u][4]=v;            dp[u][1]=cur;            dp[u][2]=ano1;        }        else if(cur>ano2)  {            dp[u][1]=ano1;            dp[u][2]=cur;        }        else{            dp[u][1]=ano1;            dp[u][2]=ano2;        }        dp[u][0]+=max(0,dp[v][0]-2*c);    }}void dfs2(int u,int f,int fback,int fnback){    ans[u]=max(dp[u][0]+fnback,dp[u][1]+fback);    for(int i=0;i<G[u].size();i++){        int v=G[u][i].v,c=G[u][i].c;        if(v==f) CT;        int uback=fback+dp[u][0]-max(0,dp[v][0]-2*c)-2*c,unback;        if(v==dp[u][4]){           unback=max(dp[u][0]-max(0,dp[v][0]-2*c)+fnback,                      fback+dp[u][2]-max(0,dp[v][0]-2*c))-c;        }        else{           unback=max(fnback+dp[u][0]-max(0,dp[v][0]-2*c),                      fback+dp[u][1]-max(0,dp[v][0]-2*c))-c;        }        dfs2(v,u,max(0,uback),max(0,unback));    }}int main(){    int cas,kk=0;    SC("%d",&cas);    while(cas--){        int n;SC("%d",&n);        MM(dp,0);        for(int i=1;i<=n;i++){            SC("%d",&val[i]);            G[i].clear();        }        for(int i=1;i<=n-1;i++){            int u,v,c;            SC("%d%d%d",&u,&v,&c);            G[u].push_back((edge){v,c});            G[v].push_back((edge){u,c});        }        dfs1(1,-1);        dfs2(1,-1,0,0);        printf("Case #%d:\n",++kk);        for(int i=1;i<=n;i++) printf("%d\n",ans[i]);    }    return 0;}

　题解：

hdu 5834 Magic boy Bi Luo with his excited tree 树形dp+转移


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