Thursday, June 9, 2016

water flow days – Moonstone


water flow days – Moonstone
题目就是给你一个matrix,里面的数字代表bar的高度,现在说降雨量如果高于bar的高度水可以漫过去,降雨量0开始每天+1这样,问最早第几天水可以有一条路径从src漫到dst

这轮也是讨论optimization很久,最后用bfs写一个subproblem。

http://www.1point3acres.com/bbs/thread-191771-1-1.html

第三题是不是就是lc 174. Dungeon Game的变种?DP从dst开始做?

这道题常规解法是O(max(day)*n^2) 当mat里有个别bar相当高的时候那么day by day增长的模式会造成大量空转 所以一个办法来避免是先把所有bar寻到一个数组并排好序 然后每当bfs到某几个bar不能再move的时候 我们从那个数组里找到某天天数大于当前bound某个bar 然后继续bfs到不能再走 这样一直到找到终点 这样总能最早到达目的地因为每次bfs完我们都是从那个数组里找到下一个最早能继续move的时间。

不是给了src和destination吗?楼主的意思是如果bfs到某个点走不动了,直接跳到下一个点继续bfs吗?这样可以吗?
我的想法是贪心+BFS,一旦有路就走到下面去,没有就选四个里面最小的那个走下去
是的 就是从那个排好序的数字搜索下一个直接跳 其实找下一个的这个时间复杂度我感觉是降不下来的 好像面试官也这么认为..

先假设这个问题是一维的,那么从src能够淹没到dst的日期取决于他俩之间最高的bar的高度。
转换为一个matrix的话,我们其实就是要找一条从src到dst的路径,这条路径上最高bar的高度越小越好。然后最短日期其实就是这条路径上最高bar的高度+1。

我们可以用BFS,然后把一般BFS其中的queue换成priority queue,在priority queue顶端的是当前未走过的高度最小的bar的index。
int leastDays(const pair<int, int>& src, const pair<int, int>& dst, const vector<vector<int>>& matrix) const {
        const vector<pair<int, int>> directions{ { 0, 1 }, { 0, -1 }, { 1, 0 }, { -1, 0 } };
        int n = matrix.size();
        if (n == 0) {
                return 0;
        }
        int m = matrix.size();
        if (m == 0) {
                return 0;
        }
        int highest = 0;
        auto comp = [&](const node& rhs, const node& lhs){ return rhs.value != lhs.value ? rhs.value > lhs.value : rhs.index > lhs.index; };
        unordered_set<pair<int, int>, MyHash> visited;
        priority_queue<node, vector<node>, decltype(comp)> q;

        visited.insert(src);
        q.emplace(src, matrix[src.first][src.second]);
        while (!q.empty()) {
                auto p = q.top();
                q.pop();
                if (p.value > highest) {
                        highest = p.value;
                }
                for (const auto& d : directions) {
                        pair<int, int> next(p.index.first + d.first, p.index.second + d.second);
                        if (next.first >= 0 && next.first < n && next.second >= 0 && next.second < m && visited.find(next) == visited.end()) {
                                if (next == dst) {
                                        highest = max(highest, matrix[dst.first][dst.second]);
                                        return highest + 1;
                                }
                                int next_value = matrix[next.first][next.second];
                                q.emplace(next, next_value);
                                visited.insert(next);
                        }
                }
        }
        return highest + 1;
}

struct MyHash {
        inline std::size_t operator()(const std::pair<int, int>& p) const {
                return (hash<int>()(p.first) + hash<int>()(p.second));
        }
};

struct node {
        pair<int, int> index;
        int value;
        node(pair<int, int> idx, int val) : index(idx), value(val) {}
};
};


    public void TestMinDaysOfWaterFlow()
    {
        int[,] matrix1 = new int[,]
        {
            {5, 8, 15, 20},
            {13, 7, 3, 1},
            {9, 10, 8, 4}
        };
        int r1 = MinDaysOfWaterFlow(matrix1, 0, 0, 2, 3);
    }
    public int MinDaysOfWaterFlow(int[,] matrix, int startRow, int startColumn, int endRow, int endColumn)
    {
        int result = int.MaxValue;
        Stack<Tuple<int, int>> stack = new Stack<Tuple<int, int>>();
        HashSet<int> visited = new HashSet<int>();
        MinDaysOfWaterFlowHelper(matrix, ref result, ref matrix[startRow, startColumn], startRow, startColumn, endRow, endColumn, visited, stack);
        return result;
    }
    private void MinDaysOfWaterFlowHelper(int[,] matrix, ref int result, ref int curMax, int startRow, int startColumn, int endRow, int endColumn, HashSet<int> visited, Stack<Tuple<int, int>> stack)
    {
        int R = matrix.GetLength(0);
        int C = matrix.GetLength(1);
        if ((startRow == endRow) && (startColumn == endColumn))
        {
            if ((result == int.MaxValue) || (curMax < result))
            {
                result = curMax;
            }
            return;
        }
        int index = startRow * R + startColumn;
        visited.Add(index);
        int[] dirX = new int[4] { 0, -1, 0, 1 };
        int[] dirY = new int[4] { -1, 0, 1, 0 };
        for (int i = 0; i < 4; i++)
        {
            int x = startRow + dirX[i];
            int y = startColumn + dirY[i];
            if ((x >= 0) && (x < R) && (y >= 0) && (y < C) && (!visited.Contains(x * R + y)))
            {
                if ((result != int.MaxValue) && (matrix[x, y] >= result))
                {
                    continue;
                }
                int preCurMax = curMax;
                if (matrix[x, y] >= curMax)
                {
                    curMax = matrix[x, y];
                }
                stack.Push(new Tuple<int, int>(x, y));
                MinDaysOfWaterFlowHelper(matrix, ref result, ref curMax, x, y, endRow, endColumn, visited, stack);
                stack.Pop();
                if (matrix[x, y] == curMax)
                {
                    curMax = preCurMax;
                }
            }
        }
        visited.Remove(index);
    }
}
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