Ural 1018 Binary Apple Tree

题目链接：http://acm.timus.ru/problem.aspx?space=1&num=1018

Dynamic Programming. 首先要根据input建立树形结构，然后在用DP来寻找最佳结果。dp[i][j]表示node i的子树上最多保存j个分支的最佳结果。代码如下：

 #include <iostream>
 #include <math.h>
 #include <stdio.h>
 #include <cstdio>
 #include <algorithm>
 #include <string.h>
 #include <string>
 #include <sstream>
 #include <cstring>
 #include <queue>
 #include <vector>
 #include <functional>
 #include <cmath>
 #include <set>
 #define SCF(a) scanf("%d", &a)
 #define IN(a) cin>>a
 #define FOR(i, a, b) for(int i=a;i<b;i++)
 #define Infinity 9999999
 typedef long long Int;
 using namespace std;
 
 int dp[][]; //row -> node id, col -> branches preserved.
 
 struct link {
     int parent;
     int child;
     int weight;
 };
 
 struct tree {
     int parent;
     tree *left, *right;
     int leftw, rightw;
 };
 
 int N, Q;
 link links[];
 bool visited[];
 
 tree * construct(int pid)
 {
     tree *t = new tree;
     t->parent = pid;
     int num = ;
     bool found = false;
     FOR(i, , N - )
     {
         if (visited[i] == false && (links[i].parent == pid || links[i].child==pid))
         {
             visited[i] = true;
             if (num == )
             {
                 t->leftw = links[i].weight;
                 if(links[i].parent==pid)
                     t->left = construct(links[i].child);
                 else
                     t->left = construct(links[i].parent);
                 num++;
             }
             else if (num == )
             {
                 t->rightw = links[i].weight;
                 if(links[i].parent==pid)
                     t->right = construct(links[i].child);
                 else
                     t->right = construct(links[i].parent);
                 found = true;
                 break;
             }
         }
     }
     if (found == false)
     {
         t->left = NULL;
         t->right = NULL;
     }
     return t;
 }
 
 int calcuTree(tree *t, int R)
 {
     if (t == NULL)
         return ;
 
     int id = t->parent;
     if (dp[id][R] == -)
     {
         int maxLeft = ;
         for (int leftR = ; leftR <= R; leftR++)
         {
             int cleft = ;
             int rightR = R - leftR;
             if (leftR > )
             {
                 cleft += t->leftw;
                 cleft += calcuTree(t->left, leftR - );
             }
             if (rightR > )
             {
                 cleft += t->rightw;
                 cleft += calcuTree(t->right, rightR - );
             }
 
             if (cleft > maxLeft)
                 maxLeft = cleft;
         }
         dp[id][R] = maxLeft;
     }
     return dp[id][R];
 }
 
 int main()
 {
     while (scanf("%d %d\n", &N, &Q) != EOF)
     {
         FOR(i, , N - )
         {
             int p, c, w;
             scanf("%d %d %d", &links[i].parent, &links[i].child, &links[i].weight);
         }
 
         FOR(i, , N - )
             visited[i] = false;
 
         tree *t = new tree;
         t = construct();
 
         FOR(i, , N + )
         {
             FOR(j, , Q + )
             {
                 dp[i][j] = -;
             }
         }
 
         FOR(i, , N + )
         {
             dp[i][] = ;
         }
 
         int ans = calcuTree(t, Q);
         printf("%d\n", ans);
     }
     return ;
 }