力扣2503.矩阵查询可获得的最大分数
-
离线算法 + 排序 + 小根堆
- 将query数组从小到大排序，用小根堆存每个单元格的值
- 从小到大遍历query，如果堆顶元素小于query[i]，弹出，直到全部弹出，总弹出个数为答案
cpp
class Solution {
int dx[4] = {0,1,0,-1},dy[4] = {1,0,-1,0};
public:
vector<int> maxPoints(vector<vector<int>>& grid, vector<int>& queries) {
int k = queries.size(),id[k];
iota(id,id+k,0);
sort(id,id+k,[&](int i,int j){return queries[i] < queries[j];});
vector<int> ans(k);
priority_queue<tuple<int, int, int>, vector<tuple<int, int, int>>, greater<>> pq;
pq.emplace(grid[0][0],0,0);
grid[0][0] = 0;
int m = grid.size(),n = grid[0].size(),cnt=0;
for(int qi:id)
{
//堆顶元素更小
int q = queries[qi];
//tuple的语法，get<0>(tuple) 返回tuple的第一个元素的引用
while(!pq.empty() && get<0>(pq.top()) < q)
{
cnt ++;
auto[_,tx,ty] = pq.top();
pq.pop();
for(int i=0;i<4;i++)
{
int x = tx + dx[i],y = ty + dy[i];
if (0 <= x && x < m && 0 <= y && y < n && grid[x][y])
{
pq.emplace(grid[x][y],x,y);
grid[x][y] = 0;
}
}
}
ans[qi] = cnt;
}
return ans;
}
};