HDU 1520 Anniversary party (树形DP)

题意：题目给出一棵树，每个节点都有其权值。如果选择了一个节点则不可以选择其父节点，问能取得的最大值。

析：一个简单的树形DP，dp[i][0] 表示结点 i不选，dp[i][1] 表示 结点 i 选，最后选最大值就好。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 6e3 + 5;
const LL mod = 1e9 + 7;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
inline int gcd(int a, int b){  return b == 0 ? a : gcd(b, a%b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
vector<int> G[maxn];
bool in[maxn];
int dp[maxn][2];

void dfs(int u){
    for(int i = 0; i < G[u].size(); ++i){
        int v = G[u][i];
        dfs(v);
        dp[u][0] += Max(dp[v][1], dp[v][0]);
        dp[u][1] += dp[v][0];
    }
    return ;
}

int main(){
    while(scanf("%d", &n) == 1){
        memset(dp, 0, sizeof dp);
        for(int i = 1; i <= n; ++i){
            scanf("%d", &dp[i][1]);
            G[i].clear();
        }
        int u, v;
        memset(in, false, sizeof in);
        while(scanf("%d %d", &u, &v) == 2 && u+v){
            G[v].push_back(u);
            in[u] = true;
        }
        int ans;
        for(int i = 1; i <= n; ++i) if(!in[i]){
            dfs(i);  ans = Max(dp[i][0], dp[i][1]);  break;
        }

        printf("%d\n", ans);
    }
    return 0;
}