hdu5834 Magic boy Bi Luo with his excited tree(树形dp)
发布时间:2021-01-25 14:26:08 所属栏目:大数据 来源:网络整理
导读: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): 723????Accepted Submission(s): 192 Problem Description ? Bi Luo is a magic boy,he also has a
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? #pragma comment(linker,"/STACK:102400000,102400000")
#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 1e9;
const int MOD = 1e9+7;
#define ll long long
#define CL(a,b) memset(a,b,sizeof(a))
#define lson (i<<1)
#define rson ((i<<1)|1)
#define N 100010
int gcd(int a,int b){return b?gcd(b,a%b):a;}
struct node
{
int v,c,next;
}e[N<<1];
int tot,head[N];
void add(int u,int v,int c)
{
e[tot].v = v;
e[tot].c = c;
e[tot].next = head[u];
head[u] = tot++;
}
int val[N];
int d_down[2][N],d_up[2][N];
///dp_down[0/1][u]:u结点往其为根的子树走,并且不走回来/走回来,能得到的最大权值
///dp_up[0/1][u]:u结点往其父亲向上走,并且不走回来/走回来,能得到的最大权值
void dfs1(int u,int fa)
{
d_down[0][u] = d_down[1][u] = val[u];
for(int i=head[u]; i!=-1; i=e[i].next)
{
int v = e[i].v;
if(v == fa) continue;
dfs1(v,u);
if(d_down[0][v]-2*e[i].c>0)
d_down[0][u] += d_down[0][v]-2*e[i].c;
}
int mx = 0;
for(int i=head[u]; i!=-1; i=e[i].next)
{
int v = e[i].v;
if(v == fa) continue;
if(d_down[0][v]-2*e[i].c>0)
mx = max(mx,(d_down[1][v]-e[i].c)-(d_down[0][v]-2*e[i].c));
else mx = max(mx,d_down[1][v]-e[i].c);
}
d_down[1][u] = d_down[0][u] + mx;
}
void dfs2(int u,int fa)
{
int mx1=0,mx2=0,tmp;
for(int i=head[u]; i!=-1; i=e[i].next)
{
int v = e[i].v;
if(v == fa) continue;
if(d_down[0][v]-2*e[i].c>0)
tmp=(d_down[1][v]-e[i].c)-(d_down[0][v]-2*e[i].c);
else tmp=d_down[1][v]-e[i].c;
if(mx1<tmp) mx2=mx1,mx1=tmp;
else if(mx2<tmp) mx2=tmp;
}
for(int i=head[u]; i!=-1; i=e[i].next)
{
int v = e[i].v;
if(v == fa) continue;
int tmp2;
if(d_down[0][v]-2*e[i].c>0)
tmp2=d_down[0][u]-(d_down[0][v]-2*e[i].c);
else tmp2=d_down[0][u];
int mx=max(d_up[0][u]-2*e[i].c,tmp2-2*e[i].c);
mx = max(mx,d_up[0][u]+tmp2-2*e[i].c-val[u]);
d_up[0][v] = val[v]+max(0,mx);
if(d_down[0][v]-2*e[i].c>0)
{
if(mx1==(d_down[1][v]-e[i].c)-(d_down[0][v]-2*e[i].c))
tmp = d_down[1][u]-(d_down[1][v]-e[i].c)+mx2;
else tmp = d_down[1][u]-(d_down[0][v]-2*e[i].c);
}
else if(d_down[1][v]-e[i].c>0)
{
if(mx1==d_down[1][v]-e[i].c)
tmp = d_down[1][u]-(d_down[1][v]-e[i].c)+mx2;
else tmp = d_down[1][u];
}
else tmp = d_down[1][u];
mx = max(d_up[1][u]-e[i].c,tmp-e[i].c);
mx = max(mx,max(d_up[0][u]+tmp-e[i].c-val[u],d_up[1][u]+tmp2-e[i].c-val[u]));
d_up[1][v] = val[v]+max(0,mx);
dfs2(v,u);
}
}
int main()
{
int T;
scanf("%d",&T);
for(int cas=1; cas<=T; cas++)
{
tot = 0;
CL(head,-1);
int n;
scanf("%d",&n);
for(int i=1; i<=n; i++)
scanf("%d",val+i);
int a,c;
for(int i=1; i<n; i++)
{
scanf("%d%d%d",&a,&b,&c);
add(a,c);
add(b,a,c);
}
dfs1(1,1);
d_up[0][1] = d_up[1][1] = val[1];
dfs2(1,1);
printf("Case #%d:n",cas);
for(int i=1; i<=n; i++)
{
printf("%dn",max(d_up[1][i]+d_down[0][i],d_up[0][i]+d_down[1][i])-val[i]);
}
}
return 0;
}
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