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;

}