LintCode 1677: Stones (Heap好题)

1677. Stones
      Given array p representing position of n stones, and array d representing distance that can be thrown.

Go from left (0 position) to right. When you encounter a stone for the kth time, throw it to right if k is odd, or skip it if k is even.

Return the position of the rightmost stone when you will not encounter a stone any more.

Example
Example 1:

Input: p = [1, 2], d = [5, 4]
Output: 11
Explanation:
First, the stone on the position 1 was thrown to 6.
Then skip the stone on the position 2.
Next the stone on the position 6 was thrown to 11.
Last skip the stone on the position 11.
Example 2:

Input: p = [1, 6], d = [5, 6]
Output: 12
Explanation:
First, the stone on the position 1 was thrown to 6.
Then skip the stone on the position 6 (the larger one).
Next the stone on the position 6 was thrown to 12.
Last skip the stone on the position 12.
Notice
n <= 10^4
p[i] <= 10^5
d[i] <= 10^3
If two or more stones stay at the same position, you will meet the largest one (the one with the smallest d[i]) first. Means throwing or skipping larger stones first.

解法1：用最小堆
代码如下。应该还可以简化一点，那个Stone里面的index好像没用。

struct Stone{
    int pos;
	int dist;
	int index;
	Stone(int p = 0, int d = 0, int id = 0) : pos(p), dist(d), index(id) {}; 
};

//min-heap
struct cmp {
    bool operator() (const Stone & a, const Stone & b) {
	    if (a.pos == b.pos) {
		    return a.dist > b.dist;
		}
		else {
		    return a.pos > b.pos;
		}
	}
};

class Solution {
public:
    /**
     * @param p: the position of the i-th stone
     * @param d: how far the stones can be stone
     * @return: the distance from the start point to the farthest stone.
     */
    int getDistance(vector<int> &p, vector<int> &d) {
        int n = p.size();
		if (n == 0) return 0;
		int result = p[0];
        
		priority_queue<Stone, vector<Stone>, cmp> pq;
		
		for (int i = 0; i < n; ++i) {
		    pq.push(Stone(p[i], d[i], i));
		}
        
		int step = 1;
		while(!pq.empty()) {
		    if (step & 0x1) {
			    Stone curStone = pq.top();
				pq.pop();
				pq.push(Stone(curStone.pos + curStone.dist, curStone.dist, curStone.index));
			} else {
			    result = pq.top().pos;
				pq.pop();
				
			}
			
		    step++;
		    
		}
		
		return result;
    }
};