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[105][105]; //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[105];
bool visited[105];

tree * construct(int pid)
{
    tree *t = new tree;
    t->parent = pid;
    int num = 0;
    bool found = false;
    FOR(i, 0, N - 1)
    {
        if (visited[i] == false && (links[i].parent == pid || links[i].child==pid))
        {
            visited[i] = true;
            if (num == 0)
            {
                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 == 1)
            {
                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 0;

    int id = t->parent;
    if (dp[id][R] == -1)
    {
        int maxLeft = 0;
        for (int leftR = 0; leftR <= R; leftR++)
        {
            int cleft = 0;
            int rightR = R - leftR;
            if (leftR > 0)
            {
                cleft += t->leftw;
                cleft += calcuTree(t->left, leftR - 1);
            }
            if (rightR > 0)
            {
                cleft += t->rightw;
                cleft += calcuTree(t->right, rightR - 1);
            }

            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, 0, N - 1)
        {
            int p, c, w;
            scanf("%d %d %d", &links[i].parent, &links[i].child, &links[i].weight);
        }

        FOR(i, 0, N - 1)
            visited[i] = false;

        tree *t = new tree;
        t = construct(1);

        FOR(i, 0, N + 1)
        {
            FOR(j, 0, Q + 1)
            {
                dp[i][j] = -1;
            }
        }

        FOR(i, 0, N + 1)
        {
            dp[i][0] = 0;
        }

        int ans = calcuTree(t, Q);
        printf("%d\n", ans);
    }
    return 0;
}