[Leetcode]64. Minimum Path Sum

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 优先队列图搜索方法，超时，查看discussion发现是因为动态规划比这个快。。 = =可能因为优先队列是一棵红黑树要插入的缘故吧。

下面是优先队列的方法。

const int x=[]{
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    return 0;
}();//这里有个链表记录它的前驱节点
struct POS{
    int x;
    int y;
    int cost;
    POS(int a, int b, int c){
        x = a;
        y = b;
        cost = c;
    }
    friend bool operator < (POS  A,POS B) {
        return A.cost > B.cost;
    } 
};
class Solution {
public:
    int minPathSum(vector<vector<int>>& grid) {
        priority_queue<POS> q;
        int m=grid.size();
        int n=grid[0].size();
        POS s(0, 0, grid[0][0]);
        q.push(s);
        while(!q.empty()){
            POS t = q.top();
            q.pop();
            if(t.x == m-1 && t.y == n-1) return t.cost;
            if(t.x+1 < m){
                POS a(t.x+1,t.y,t.cost+grid[t.x+1][t.y]);
                q.push(a);
            }
            if(t.y+1 < n){
                POS a(t.x,t.y+1,t.cost+grid[t.x][t.y+1]);
                q.push(a);
            }
        }
        return 0;
    }
};

下面是动态规划的方式

const int x=[]{
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    return 0;
}();//这里有个链表记录它的前驱节点
const int inf=0x3f3f3f;
class Solution {
public:
    int minPathSum(vector<vector<int>>& grid) {
        int m=grid.size();
        int n=grid[0].size();
        int *cost = new int[m*n];
        for(int i = 0; i < m*n; i++) cost[i]=inf;
        
        for(int i = 0; i < m ; i++){
            for(int j = 0; j < n; j++){
                if(i == 0 && j == 0) cost[i * n + j] = grid[i][j];
                if(j > 0){
                    cost[i * n + j] = cost[i * n + j] < cost[i * n + j - 1] + grid[i][j]?cost[i * n + j] : cost[i * n + j - 1] + grid[i][j];
                }
                if(i > 0){
                    cost[i * n + j] = cost[i * n + j] < cost[(i-1) * n + j] + grid[i][j]?cost[i * n + j] : cost[(i-1) * n + j] + grid[i][j];
                }
            }
        }
        int ans=cost[m*n-1];
        delete cost;
        return ans ;
    }
};