HDU 5242 Game (贪心)

题意：给定一棵树，要求从根结点1走k次，每次都是到叶子结点结束，把走过的所有的结点权值加起来，最大是多少。

析：先把每个结点到根结点的路径之和求出来，然后按权值从大到小排序，然后每次把路径中的权值求出来，最后求前k个值的和即可。

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 <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
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 bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}
struct Node{
  int v;
  LL val;
  bool operator < (const Node &p) const{
    return val > p.val;
  }
};
vector<Node> v;
int p[maxn], a[maxn];
LL dp[maxn];
vector<int> G[maxn];

void dfs(int u){
  for(int i = 0; i < G[u].size(); ++i){
    int vv = G[u][i];
    dp[vv] = dp[u] + a[vv];
    if(G[vv].size() == 0)  v.push_back((Node){vv, dp[vv]});
    dfs(vv);
    p[vv] = u;
  }
}
bool vis[maxn];

int main(){
  int T;  cin >> T;
  for(int kase = 1; kase <= T; ++kase){
    scanf("%d %d", &n, &m);
    for(int i = 1; i <= n; ++i){
      scanf("%d", a+i);
      G[i].clear();
    }
    for(int i = 1; i < n; ++i){
      int u, v;
      scanf("%d %d", &u, &v);
      G[u].push_back(v);
    }
    dp[1] = a[1];
    v.clear();
    p[1] = 0;
    dfs(1);
    sort(v.begin(), v.end());
    vector<LL> tmp;
    memset(vis, 0, sizeof vis);
    for(int i = 0; i < v.size(); ++i){
      Node& u = v[i];
      int x = u.v;
      while(x && !vis[x]){
        vis[x] = 1;
        x = p[x];
      }
      tmp.push_back(u.val-dp[x]);
    }
    sort(tmp.begin(), tmp.end(), greater<LL>());
    LL ans = 0;
    int t = min((int)tmp.size(), m);
    for(int i = 0; i < t; ++i)  ans += tmp[i];
    printf("Case #%d: %lld
", kase, ans);
  }
  return 0;
}