HDU 4118 Holiday's Accommodation (dfs)

题意：给n个点，每个点有一个人，有n-1条有权值的边，求所有人不在原来位置所移动的距离的和最大值。

析：对于每边条，我们可以这么考虑，它的左右两边的点数最少的就是要加的数目，因为最好的情况就是左边到右边，右边到左边，然后用dfs就可以解决了。

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>
#include <list>
#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 double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 10;
const int mod = 1000;
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, val, next;
};
Node a[maxn<<1];
int head[maxn];
int cnt;

void add(int u, int v, int val){
  a[cnt].v = v;
  a[cnt].val = val;
  a[cnt].next = head[u];
  head[u] = cnt++;
}
LL ans;
int num[maxn];
void dfs(int u, int fa){
  for(int i = head[u]; ~i; i = a[i].next){
    int v = a[i].v;
    if(v == fa)  continue;
    dfs(v, u);
    num[u] += num[v];
    ans += (LL)a[i].val * min(num[v], n-num[v]);
  }
  ++num[u];
}

int main(){
  int T;  cin >> T;
  for(int kase = 1; kase <= T; ++kase){
    scanf("%d", &n);
    memset(head, -1, sizeof head);
    cnt = 0;
    for(int i = 1; i < n; ++i){
      int u, v, val;
      scanf("%d %d %d", &u, &v, &val);
      add(u, v, val);
      add(v, u, val);
    }
    ans = 0;
    memset(num, 0, sizeof num);
    dfs(1, -1);
    printf("Case #%d: %I64d\n", kase, ans*2LL);
  }
  return 0;
}