树型dp hdu5834 Magic boy Bi Luo with his excited tree

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题意：一棵树，对于每个点出发，结束位置可以是任意的，走过的点权值只加一次，走过的边权值要减去走过的次数乘以边权值。

问对于每一个点，权值和最大是多少。

思路：

我们需要维护4个内容

A[u]表示从u往下走，并回到u，路上的最大权值之和

B[u]表示从u往下走，不需要回到u，路上的最大权值之和

C[u]表示从u往上走，需要回到u，路上的最大权值之和

D[u]表示从u往上走，不需要回到u，路上的最大权值之和

上面这4种权值之和都不包括u本身

W[u]表示u节点的权值

f(x)=max(x,0)

若v是u的第i个节点，第一遍DFS从下往上更新,那么有

pre[i]=pre[i−1]+f(A[v]+W[v]−2∗cost)

suf[i]=suf[i+1]+f(A[v]+W[v]−2∗cost)

A[u]=suf[1]

B[u]=max(pre[i−1]+suf[i+1]+f(B[v]+W[v]−cost))

第二遍DFS从上往下更新，有

suf2[i]=max(suf2[i−1],f(B[v]+W[v]−cost)−f(A[v]+W[v]−2∗cost))

pre2[i]=max(pre2[i−1],f(B[v]+W[v]−cost)−f(A[v]+W[v]−2∗cost))

C[v]=f(pre[i−1]+suf[i+1]+C[u]+W[u]−2∗cost)

D[v]=f(pre[i−1]+suf[i+1]+C[u]+W[u]+max(pre2[i−1],suf2[i+1])−cost)

D[v]=max(D[v],f(pre[i−1]+suf[i+1]+D[u]+W[u]−cost)

答案ans[u]=max(A[u]+D[u],B[u]+C[u])+W[u]

#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <stack>
#include <queue>
#include <cstdio>
#include <cctype>
#include <bitset>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
#define fuck(x) cout<<"["<<x<<"]";
#define FIN freopen("input.txt","r",stdin);
#define FOUT freopen("output.txt","w+",stdout);
//#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
typedef long long LL;
typedef pair<int, int>PII;

const int MX = 1e5 + 5;
const int INF = 0x3f3f3f3f;

int Head[MX], erear;
struct Edge {
int v, cost, nxt;
} E[MX * 2];
void edge_init() {
erear = 0;
memset(Head, -1, sizeof(Head));
}
void edge_add(int u, int v, int cost) {
E[erear].v = v;
E[erear].cost = cost;
E[erear].nxt = Head[u];
Head[u] = erear++;
}

int ans[MX];
int W[MX], A[MX], B[MX], C[MX], D[MX];
int get_child(int u, int f, vector<int> &ch) {
int tot = 0;
for(int i = Head[u]; ~i; i = E[i].nxt) {
int v = E[i].v;
if(v == f) continue;
tot++;
}
ch.resize(tot + 1); tot = 0;
for(int i = Head[u]; ~i; i = E[i].nxt) {
int v = E[i].v;
if(v == f) continue;
ch[++tot] = i;
}
return tot;
}
inline int func(int x) {
return max(x, 0);
}
void DFS1(int u, int f) {
vector<int> ch, suf;
int tot = get_child(u, f, ch);
suf.resize(tot + 2);

A[u] = B[u] = 0;
suf[tot + 1] = 0;
for(int i = tot; i >= 1; i--) {
int v = E[ch[i]].v, cost = E[ch[i]].cost;
DFS1(v, u);
suf[i] = suf[i + 1] + func(A[v] + W[v] - 2 * cost);
}
A[u] = suf[1];

int pre = 0;
for(int i = 1; i <= tot; i++) {
int v = E[ch[i]].v, cost = E[ch[i]].cost;
B[u] = max(B[u], pre + suf[i + 1] + func(B[v] + W[v] - cost));
pre += func(A[v] + W[v] - 2 * cost);
}
}
void DFS2(int u, int f) {
ans[u] = max(A[u] + D[u], B[u] + C[u]) + W[u];

vector<int> ch, suf, suf2;
int tot = get_child(u, f, ch);
suf.resize(tot + 2);
suf2.resize(tot + 2);

suf[tot + 1] = suf2[tot + 1] = 0;
for(int i = tot; i >= 1; i--) {
int v = E[ch[i]].v, cost = E[ch[i]].cost;
suf[i] = suf[i + 1] + func(A[v] + W[v] - 2 * cost);
suf2[i] = max(suf2[i + 1], func(B[v] + W[v] - cost) - func(A[v] + W[v] - 2 * cost));
}

int pre = 0, pre2 = 0;
for(int i = 1; i <= tot; i++) {
int v = E[ch[i]].v, cost = E[ch[i]].cost;
C[v] = func(pre + suf[i + 1] + C[u] + W[u] - 2 * cost);
D[v] = func(pre + suf[i + 1] + C[u] + W[u] + max(pre2, suf2[i + 1]) - cost);
D[v] = max(D[v], func(pre + suf[i + 1] + D[u] + W[u] - cost));
pre += func(A[v] + W[v] - 2 * cost);
pre2 = max(pre2, func(B[v] + W[v] - cost) - func(A[v] + W[v] - 2 * cost));
DFS2(v, u);
}
}

int main() {
//FIN;
int T, n, ansk = 0;
scanf("%d", &T);
while(T--) {
printf("Case #%d:\n", ++ansk);
edge_init();
scanf("%d", &n);
for(int i = 1; i <= n; i++) scanf("%d", &W[i]);
for(int i = 1; i <= n - 1; i++) {
int u, v, cost;
scanf("%d%d%d", &u, &v, &cost);
edge_add(u, v, cost);
edge_add(v, u, cost);
}
DFS1(1, -1);

C[1] = D[1] = 0;
DFS2(1, -1);
for(int i = 1; i <= n; i++) {
printf("%d\n", ans[i]);
// printf("%d,%d,%d,%d\n", A[i], B[i], C[i], D[i]);
}
}
return 0;
}