tjut 4616

来源：互联网发布：mysql 登录失败处理编辑：程序博客网时间：2024/06/06 16:02
#pragma comment(linker,"/STACK:102400000,102400000")  #include <iostream>  #include <cstring>  #include <cstdio>  #include <vector>  #include <cmath>  using namespace std;  #define ReadFile(str) freopen(str,"r",stdin);  #define CLR(c,v) memset(c,v,sizeof(c))  typedef long long ll;  const int N = 5e4 + 5;    struct Node{      int val;  // 权值      int tra;  // 陷阱      bool vis; // 判断回头路      ll dp[5][2];  }node[N];  vector <int > tree[N];  ll ans;  int n,c;    void dfs(int father){ // father,child(id)  nfather,nchild(node)       int size = tree[father].size();      node[father].vis = true;      Node& nfather = node[father];            ///// init dp      int& trap = nfather.tra;      nfather.dp[trap][0] = nfather.dp[trap][1] = nfather.val;        for(int i = 0 ; i < size ; i++){          int  child  = tree[father][i];          Node& nchild = node[child];          if(nchild.vis)continue;          dfs(child);          for(int j = 0 ; j <= c ; j++){              for(int k = 0 ; k+j <= c ; k++){                  if(j != c) // 若j==c了 那么在父节点得到的最大值就没有机会再往孩子下面走了 否则还有机会往下走                      ans = max(ans , nfather.dp[j][0] + nchild.dp[k][1]); //所以转移是到父节点+另一个孩子回来                  if(k != c) // 同理                      ans = max(ans , nfather.dp[j][1] + nchild.dp[k][0]);                   if(j + k < c)// 以nfather为衔接点 当i+k==c的时候有(1||2)个端点一定在陷阱上                      ans = max(ans , nfather.dp[j][0] + nchild.dp[k][0]);                                         //ans = max(ans , nfather.dp[j][1] + nchild.dp[k][1]);               }          }          for(int j = 0 ; j <= c ; j++){              if(j+trap>c)break;              nfather.dp[j+trap][0] = max(nfather.dp[j+trap][0] , nfather.val + nchild.dp[j][0]);                  if(j) // 当前机会数 必须大于0才能网上推              nfather.dp[j+trap][1] = max(nfather.dp[j+trap][1] , nfather.val + nchild.dp[j][1]);              }      }  }    int main(){      //ReadFile("in.txt");      int T;cin >> T;      while(T--){          cin >> n >> c;          for(int i = 0 ; i < n ; i++){              scanf("%d %d", &node[i].val, &node[i].tra);              node[i].vis = false;              CLR(node[i].dp,0);          }          for(int i = 1 ; i < n ; i++){              int u,v;              scanf("%d %d", &u, &v);              tree[u].push_back(v);              tree[v].push_back(u);          }          ans = 0;          dfs(0);          cout << ans << endl;          for(int i = 0; i <= n ; i++)              tree[i].clear();      }      return 0;  }
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