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HackerRank - "Coin on the Table" - 3D DP

时间:2015-05-15 17:14:33      阅读:114      评论:0      收藏:0      [点我收藏+]

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This is my 1st 3D DP problem, and it should be an educational one. But, still tricky to me.
Here is a good article: http://www.cnblogs.com/sunshineatnoon/p/3919423.html, and I simply re-code it in C++, for learning :)

#include <cmath>
#include <cstdio>
#include <vector>
#include <map>
#include <set>
#include <unordered_set>
#include <string>
#include <climits>
#include <iostream>
#include <algorithm>
#include <unordered_map>
#include <unordered_set>
using namespace std;

int CalcMin(int i, int j, int k,
    vector<vector<vector<int>>> &dp,
    vector<string> &map)
{
    int min_mv = INT_MAX - 1;
    int n = map.size();
    int m = map[0].size();

    if (i - 1 >= 0)
    {
        int newMv = dp[i - 1][j][k - 1] + (map[i - 1][j] == D ? 0 : 1);
        min_mv = std::min(min_mv, newMv);
    }
    if (i + 1 < n)
    {
        int newMv = dp[i + 1][j][k - 1] + (map[i + 1][j] == U ? 0 : 1);
        min_mv = std::min(min_mv, newMv);
    }
    if (j - 1 >= 0)
    {
        int newMv = dp[i][j - 1][k - 1] + (map[i][j - 1] == R ? 0 : 1);
        min_mv = std::min(min_mv, newMv);
    }
    if (j + 1 < m)
    {
        int newMv = dp[i][j + 1][k - 1] + (map[i][j + 1] == L ? 0 : 1);
        min_mv = std::min(min_mv, newMv);
    }
    return min_mv;
}

int main()
{
    int n, m, K;
    cin >> n >> m >> K;

    int sx, sy;

    vector<string> map;
    for (int i = 0; i < n; i++)
    {
        char buf[60] = { 0 };
        scanf("%s", buf);
        string s(buf);
        size_t inx;
        if ((inx = s.find(*)) != string::npos)
        {
            sx = i;
            sy = inx;
        }
        map.push_back(s);
    }

    //
    vector<vector<vector<int>>> dp(n, vector<vector<int>>(m, vector<int>(K + 1)));
    for (int k = 0; k <= K; k++)
    for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
    {
        if (k == 0)    // INIT
        {
            dp[i][j][k] = ((i + j) == 0) ? 0 : INT_MAX - 1;
        }
        else
        {
            dp[i][j][k] = CalcMin(i, j, k, dp, map);
        }
    }

    //    Get Min
    int ret = INT_MAX - 1;
    for (int k = 0; k <= K; k++)
        ret = std::min(ret, dp[sx][sy][k]);

    cout << (ret == (INT_MAX - 1) ? -1 : ret) << endl;
    return 0;
}

 

HackerRank - "Coin on the Table" - 3D DP

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原文地址:http://www.cnblogs.com/tonix/p/4506132.html

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