HDU 3586 Information Disturbing (树形DP+二分)

题意：给出n个士兵，其中1号为指挥官，关系为树状结构，叶子为先锋，现在要在总花费小于等于m的情况切断所有的先锋与指挥官的联系，问最大的限制最小为多少。

析：很明显是一个树形DP，但是限制怎么求呢，就是通过二分，然后变成一个判定性问题，dp[i] 表示切断 以 i 的子树的最少花费不多少，当然是不超过限制的，这里就状态转移方程也是好写的dp[i] += min(dp[v], val)  v 是u的子结点，val 是该边的值。

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>
#include <assert.h>
#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 fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define FOR(x,n)  for(int i = (x); i < (n); ++i)
#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 = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1000 + 5;
const LL 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 Edge{
  int to, val, next;
};

Edge edge[maxn<<1];
int head[maxn], cnt;

void addEdge(int u, int v, int val){
  edge[cnt].val = val;
  edge[cnt].to = v;
  edge[cnt].next = head[u];
  head[u] = cnt++;
}

int dp[maxn];

void dfs(int u, int fa, int mid){
  dp[u] = 0;
  bool ok = true;
  for(int i = head[u]; ~i; i = edge[i].next){
    int v = edge[i].to;
    if(v == fa)  continue;
    ok = false;
    dfs(v, u, mid);
    if(dp[u] == INF)  continue;
    if(edge[i].val <= mid)
      dp[u] += min(edge[i].val, dp[v]);
    else{
      if(dp[v] == INF)  dp[u] = INF;
      else dp[u] += dp[v];
    }
  }
  if(ok)  dp[u] = INF;
}

bool judge(int mid){
  dfs(1, -1, mid);
  return dp[1] <= m;
}


int main(){
  while(scanf("%d %d", &n, &m) == 2 && n+m){
    ms(head, -1);
    cnt = 0;
    int mmax = -1;
    for(int i = 1; i < n; ++i){
      int u, v, c;
      scanf("%d %d %d", &u, &v, &c);
      mmax = max(mmax, c);
      addEdge(u, v, c);
      addEdge(v, u, c);
    }
    int l = 1, r = mmax + 1;
    while(l < r){
      int mid = l + r >> 1;
      if(judge(mid))  r = mid;
      else l = mid + 1;
    }
    printf("%d\n", l == mmax+1 ? -1 : l);
  }
  return 0;
}