題目鏈接：

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): 1037????Accepted Submission(s): 298

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?

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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.

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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.

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Sample Input 1 5 4 1 7 7 7 1 2 6 1 3 1 2 4 8 3 5 2

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Sample Output Case #1: 15 10 14 9 15 題意： 每個節點有價值v[i]的寶物,但是任何兩個節點u,v之間的路走一次花費為w,從每個節點出發最多可以賺多少錢; 思路： 樹形dp的題目,需要記錄轉移的最大和次大,注意轉移的情況,不能寫漏了; AC代碼： #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <bits/stdc++.h> #include <stack> #include <map>using namespace std;#define For(i,j,n) for(int i=j;i<=n;i++) #define mst(ss,b) memset(ss,b,sizeof(ss));typedef long long LL;template<class T> void read(T&num) {char CH; bool F=false;for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());F && (num=-num); } int stk[70], tp; template<class T> inline void print(T p) {if(!p) { puts("0"); return; }while(p) stk[++ tp] = p%10, p/=10;while(tp) putchar(stk[tp--] + '0');putchar('\n'); }const int mod=1e9+7; const double PI=acos(-1.0); const int inf=1e9; const int N=(1<<20)+10; const int maxn=1e5+110; const double eps=1e-12;int n,cnt,head[maxn],a[maxn]; int down[maxn][2],up[maxn][2],max1[maxn],max2[maxn],temp[maxn],cost[maxn]; struct Edge {int from,to,next,val; }edge[2*maxn]; inline void add_edge(int s,int e,int va) {edge[cnt].from=s;edge[cnt].to=e;edge[cnt].next=head[s];edge[cnt].val=va;head[s]=cnt++; } inline void Init() {cnt=0;for(int i=0;i<=n;i++)head[i]=-1; } void dfs(int cur,int fa,int va) {down[cur][1]=a[cur];cost[cur]=va;for(int i=head[cur];i!=-1;i=edge[i].next){int x=edge[i].to;if(x==fa)continue;dfs(x,cur,edge[i].val);if(down[x][1]-2*edge[i].val>=0)down[cur][1]+=down[x][1]-2*edge[i].val;} } void dfs1(int cur,int fa) {down[cur][0]=a[cur];temp[cur]=max1[cur]=max2[cur]=0;for(int i=head[cur];i!=-1;i=edge[i].next){int x=edge[i].to;if(x==fa)continue;dfs1(x,cur);if(down[x][0]-edge[i].val>0){int t=down[cur][1];if(down[x][1]-2*edge[i].val>=0)t-=down[x][1]-2*edge[i].val;t+=down[x][0]-edge[i].val;if(t>=down[cur][0]){max2[cur]=max1[cur];temp[cur]=down[cur][0];down[cur][0]=t;max1[cur]=x;}else if(t>temp[cur]){max2[cur]=x;temp[cur]=t;}}} }void dfs2(int cur,int fa,int va) {up[cur][1]=0;if(down[cur][1]-2*va>=0)up[cur][1]=max(up[cur][1],down[fa][1]-down[cur][1]+2*va+up[fa][1]-2*va);else up[cur][1]=max(up[cur][1],down[fa][1]+up[fa][1]-2*va);up[cur][0]=0;if(max1[fa]==cur){int t=down[fa][0]-down[cur][0]+va;up[cur][0]=max(up[cur][0],t+up[fa][0]-va);int r=max2[fa];if(down[r][1]-2*cost[r]>0)t=t-down[r][1]+2*cost[r];t+=down[r][0]-cost[r];up[cur][0]=max(up[cur][0],t+up[fa][1]-va);}else {int t=down[fa][0];if(down[cur][1]-2*va>0)t-=down[cur][1]-2*va;up[cur][0]=max(up[cur][0],t+up[fa][1]-va);t=down[fa][1];if(down[cur][1]-2*va>0)t-=down[cur][1]-2*va;up[cur][0]=max(up[cur][0],t+up[fa][0]-va);}for(int i=head[cur];i!=-1;i=edge[i].next){int x=edge[i].to;if(x==fa)continue;dfs2(x,cur,edge[i].val);} }int main() {int t,Case=0;read(t);while(t--){read(n);Init();For(i,1,n)read(a[i]);int u,v,w;For(i,1,n-1){read(u);read(v);read(w);add_edge(u,v,w);add_edge(v,u,w);}dfs(1,0,0);dfs1(1,0);dfs2(1,0,0);printf("Case #%d:\n",++Case);for(int i=1;i<=n;i++)printf("%d\n",max(down[i][0]+up[i][1],down[i][1]+up[i][0]));} return 0; }

　　

轉載于:https://www.cnblogs.com/zhangchengc919/p/5856654.html

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