标签：create ios add 题解 div positive font ace hot
InputFirst line is a positive integer T(T≤104)T(T≤104) , represents there are TT test cases.
Four each test：
The first line contain an integer NN(N≤105)(N≤105).
The next line contains NN integers V[i]V[i], which means the treasure’s value of node i(1≤V[i]≤104)i(1≤V[i]≤104).
For the next N−1N−1 lines, each contains three integers u,v,cu,v,c , which means node uuand node vv are connected by an edge, it‘s cost is c(1≤c≤104)c(1≤c≤104).
You can assume that the sum of NN will not exceed 106106.OutputFor the i-th test case , first output Case #i: in a single line , then output NNlines , for the i-th line , output ans[i]ans[i] in a single line.Sample Input
1
5
4 1 7 7 7
1 2 6
1 3 1
2 4 8
3 5 2
Sample Output
Case #1:
15
10
14
9
15
SOLUTION:
这算是一道比较难的dp了吧，
很容易想到dp i 0/1 代表只考虑i的子树后回到，不会到根节点的最大价值，
之后就要考虑此时的转移
求出dp之后，最难的步骤就是dp数组在树上换根时候的变换，感觉网上的题解写的不错
CODE:
//
// Created by Running Photon
// Copyright (c) 2015 Running Photon. All rights reserved.
//
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <sstream>
#include <set>
#include <vector>
#include <stack>
#define ALL(x) x.begin(), x.end()
#define INS(x) inserter(x, x,begin())
#define ll long long
#define CLR(x) memset(x, 0, sizeof x)
using namespace std;
const int inf = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int maxn = 2e5 + 10;
const int maxv = 1e5 + 10;
const double eps = 1e-9;
int pnt[maxn], nxt[maxn], val[maxv], head[maxv], cost[maxn], cnt;
void add_edge(int u, int v, int value) {
pnt[cnt] = v;
cost[cnt] = value;
nxt[cnt] = head[u];
head[u] = cnt++;
}
int dp[2][maxv];
void dfs1(int u, int fa) {
dp[0][u] = val[u];
dp[1][u] = val[u];
for(int i = head[u]; ~i; i = nxt[i]) {
int v = pnt[i];
if(v == fa) continue;
dfs1(v, u);
if(dp[0][v] - cost[i] * 2 > 0) {
dp[0][u] += dp[0][v] - cost[i] * 2;
}
}
for(int i = head[u]; ~i; i = nxt[i]) {
int v = pnt[i];
if(v == fa) continue;
if(dp[0][u] - max(0, dp[0][v] - cost[i] * 2) + (dp[1][v] - cost[i]) >= dp[1][u]) {
dp[1][u] = dp[0][u] - max(0, dp[0][v] - cost[i] * 2) + (dp[1][v] - cost[i]);
}
}
}
void dfs2(int u, int fa, int Cost) {
int dir = u, tmp;
if(fa != -1 && dp[0][fa] - Cost * 2 > 0) {
dp[0][u] += dp[0][fa] - Cost * 2;
}
tmp = dp[1][u] = dp[0][u];
if(fa!=-1)
for(int i=1;i<=1;i++) {
int v = fa;
if(dp[0][u] - max(0, dp[0][v] - Cost * 2) + (dp[1][v] - Cost) >= dp[1][u]) {
dir = v;
swap(tmp, dp[1][u]);
dp[1][u] = dp[0][u] - max(0, dp[0][v] - Cost * 2) + (dp[1][v] - Cost);
}
else
if(dp[0][u] - max(0, dp[0][v] - Cost * 2) + (dp[1][v] - Cost) >= tmp) {
tmp = dp[0][u] - max(0, dp[0][v] - Cost * 2) + (dp[1][v] - Cost);
}
}
for(int i = head[u]; ~i; i = nxt[i]) {
int v = pnt[i];
if(dp[0][u] - max(0, dp[0][v] - cost[i] * 2) + (dp[1][v] - cost[i]) >= dp[1][u]) {
dir = v;
swap(tmp, dp[1][u]);
dp[1][u] = dp[0][u] - max(0, dp[0][v] - cost[i] * 2) + (dp[1][v] - cost[i]);
}
else if(dp[0][u] - max(0, dp[0][v] - cost[i] * 2) + (dp[1][v] - cost[i]) >= tmp) {
tmp = dp[0][u] - max(0, dp[0][v] - cost[i] * 2) + (dp[1][v] - cost[i]);
}
}
// printf("dp[0][%d] = %d dp[1][%d] = %d tmp = %d\n", u, dp[0][u], u, dp[1][u], tmp);
int L = dp[0][u], R = dp[1][u];
for(int i = head[u]; ~i; i = nxt[i]) {
int v = pnt[i];
if(v == fa) continue;
if(dp[0][v] - cost[i] * 2 > 0) {
dp[0][u] = L - (dp[0][v] - cost[i] * 2);
}
if(dir == v) {
dp[1][u] = tmp;
if(dp[0][v] - cost[i] * 2 > 0) dp[1][u] -= (dp[0][v] - cost[i] * 2);
}
else if(dp[0][v] - cost[i] * 2 > 0) {
dp[1][u] = R - (dp[0][v] - cost[i] * 2);
}
// printf("dir = %d\n", dir);
// printf("nxt%d = %d\n", v, dp[1][u]);
dfs2(v, u, cost[i]);
dp[0][u] = L, dp[1][u] = R;
}
}
int main() {
int T;
int cas = 0;
scanf("%d", &T);
while(T--) {
int n;
printf("Case #%d:\n", ++cas);
scanf("%d", &n);
cnt = 0;
memset(head, -1, sizeof (int) * (n + 1));
for(int i = 1; i <= n; i++) scanf("%d", val + i);
for(int i = 1; i < n; i++) {
int u, v, value;
scanf("%d%d%d", &u, &v, &value);
add_edge(u, v, value);
add_edge(v, u, value);
}
dfs1(1, -1);
dfs2(1, -1, 0);
for(int i = 1; i <= n; i++) {
printf("%d\n", max(dp[0][i], dp[1][i]));
}
}
return 0;
}
Magic boy Bi Luo with his excited tree (树形dp）
标签：create ios add 题解 div positive font ace hot
原文地址：https://www.cnblogs.com/zhangbuang/p/11371836.html
