codeforces round 646 E
題目
題目大意
有一個樹,每一個節點有一個數字0或着1,要改變一個節點或者其子樹的值,改動一個需要ai的代價,問改稱目的狀態需要最小的代價是多少
思路
首先,我們可以貪心的求一下,每一個節點最小改動價值,若節點i的父親是j,那麼對於i節點來說,a[i] = min (a[i], a[j]), dfs改動一輪之後,記錄每一個點的0->1和1->0的有多少對,改動一下記錄價值,回溯一下就行了
代碼
#include <iostream>
#include <cstdio>
#include <set>
#include <list>
#include <vector>
#include <stack>
#include <queue>
#include <map>
#include <string>
#include <sstream>
#include <algorithm>
#include <cstring>
#include <cstdlib>
#include <cctype>
#include <cmath>
#include <fstream>
#include <iomanip>
//#include <unordered_map>
using namespace std;
#define dbg(x) cerr << #x " = " << x << endl;
typedef pair<int, int> P;
typedef long long ll;
#define FIN freopen("in.txt", "r", stdin);
const int MAXN = 2e5 + 5;
vector<int> v[MAXN];
ll a[MAXN], b[MAXN], c[MAXN];
int vis[MAXN];
ll ans = 0;
P dfs(int u, int fa, ll w)
{
P aa = P(0 ,0);
if(b[u] != c[u])
{
if(b[u])
{
aa.first++;
}
else
{
aa.second++;
}
}
for (int i = 0; i < v[u].size(); i++)
{
int next = v[u][i];
if(next == fa) continue;
P tmp = dfs(next, u, min(a[u], w));
aa.first += tmp.first;
aa.second += tmp.second;
}
if(a[u] < w)
{
ll take = min(aa.first, aa.second);
ans += take * a[u] * 2;
aa.first -= take;
aa.second -= take;
}
return aa;
}
int main()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n;
cin >> n;
int flg = 0;
for (int i = 1; i <= n; i++)
{
cin >> a[i] >> b[i] >> c[i];
flg += b[i] - c[i];
}
for (int i = 0; i < n - 1; i++)
{
int st, ed;
cin >> st >> ed;
v[st].push_back(ed);
v[ed].push_back(st);
}
if (flg)
cout << "-1" << endl;
else
{
dfs(1, 0, 2e9);
cout << ans << endl;
}
return 0;
}