sicily 1158. Pick numbers bfs

//这题一开始感觉是dp,状态转移方程为dp[i][j] = min(dp[i][j-1], dp[i-1][j]) + maze[i][j]

//提交上去WA

//后来发现这题不能用dp,因为题目要求的是正的最小值，状态转移方程的min会取到负数，

//当求出dp[n][m]为负数时，有可能存在一条和为正的路径

//注意到n,m都比较小，可以直接用bfs搜索过

#include <iostream>

#include <queue>

using namespace std;



const int N = 15;

const int INF = 1000000000;



int maze[N][N];

int n, m;

int ans;

int dx[] = {1, 0};

int dy[] = {0, 1};



struct status

{

	int row,col,sum;

	status() {}

	status (int row, int col, int sum)

	{

		this->row = row;

		this->col = col;

		this->sum = sum;

	}

};



void bfs()

{

	queue<status> Q;

	Q.push(status(1, 1, maze[1][1]));

	status p,tmp;



	while (!Q.empty())

	{

		p = Q.front();

		Q.pop();

		if (p.row == n && p.col == m && p.sum > 0)

			ans = min(p.sum, ans);

		

		for (int i = 0; i < 2; i++)

		{

			tmp = p;

			tmp.row += dx[i];

			tmp.col += dy[i];

			if (tmp.row >= 1 && tmp.row <= n && tmp.col >= 1 && tmp.col <= m)

			{

				tmp.sum += maze[tmp.row][tmp.col];

				Q.push(tmp);

			}

		}

	}

}



int main()

{

	while (cin >> n >> m)

	{

		for (int i = 1; i <= n; i++)

			for (int j = 1; j <= m; j++)

				cin >> maze[i][j];



		ans = INF;

		bfs();

		if (ans < INF)

			cout << ans << endl;

		else

			cout << -1 << endl;	

	}

	return 0;

}