HDU 4340 树形DP

 1 #include <map>
 2 #include <set>
 3 #include <cmath>
 4 #include <ctime>
 5 #include <queue>
 6 #include <stack>
 7 #include <cstdio>
 8 #include <string>
 9 #include <vector>
10 #include <cstring>
11 #include <sstream>
12 #include <iostream>
13 #include <algorithm>
14 #include <bitset>
15 #include <climits>
16 using namespace std;
17 
18 #define wh while
19 #define inf (int)(~0u/2)
20 #define FOR(i, n) for(int i = 0; i < n; i++)
21 #define FOR1(i, n) for(int i = 1; i < n; i++)
22 #define FOR2(i, n) for(int i = 0; i <= n; i++)
23 #define REP(i,n) for(int i = 1; i <= n; i++)
24 #define FORI(it,n) for(typeof(n.begin()) it = n.begin(); it != n.end(); it++)
25 #define sf scanf
26 #define pf printf
27 #define frs first
28 #define sec second
29 #define psh push_back
30 #define mkp make_pair
31 #define PB(x) push_back(x)
32 #define MP(x, y) make_pair(x, y)
33 #define clr(abc,z) memset(abc,z,sizeof(abc))
34 #define lt(v) v << 1
35 #define rt(v) v << 1 | 1
36 //#define mid ((l + r) >> 1)
37 #define lson l, mid, v << 1
38 #define rson mid + 1, r, v << 1 | 1
39 
40 #define fre freopen("1.txt", "r", stdin)
41 
42 typedef long long LL;
43 typedef long double LD;
44 const int INF = 0x3F3F3F3F;
45 const int maxn = 120;
46 vector<int> G[maxn];
47 int dp[maxn][2][2];
48 int cost[maxn][2]; int N;
49 
50 void dfs(int u, int fa){
51     if(G[u].size() == 1 && fa != 0){
52         FOR(i, 2){
53             dp[u][i][0] = cost[u][i] / 2;
54             dp[u][i][1] = cost[u][i];
55         }
56         return ;
57     }
58     FOR(i, (int)G[u].size()){
59         int v = G[u][i];
60         if(v == fa) continue;
61         dfs(v, u);
62     }
63     FOR(i, 2){
64         int sum = 0, cha = INF;
65         int cst = cost[u][i];
66         FOR(j, (int)G[u].size()){
67             int v = G[u][j];
68             if(v == fa) continue;
69             int tmp = min(dp[v][i][0], dp[v][i ^ 1][1]); sum += tmp;
70             cha = min(cha, dp[v][i][1] - tmp);
71             //将原先选中的dp[v][i][0]或dp[v][i ^ 1][1]替换成dp[v][i][1]所需要的最小代价
72         }
73         dp[u][i][0] = cst / 2 + sum;
74         dp[u][i][1] = min(cst + sum, cst / 2 + sum + cha);
75     }
76 }
77 
78 int main() {
79     wh(sf("%d", &N) != EOF){
80         REP(i, N) G[i].clear();
81         REP(i, N) sf("%d", &cost[i][0]);
82         REP(i, N) sf("%d", &cost[i][1]);
83         REP(i, N - 1){
84             int u, v; sf("%d%d", &u, &v);
85             G[u].PB(v); G[v].PB(u);
86         }
87         dfs(1, 0);
88         pf("%d\n", min(dp[1][0][1], dp[1][1][1]));
89     }
90 }