Grid

You are on an nxm grid where each square on the grid has a digit on it. From a given square that has digit k on it, a Move consists of jumping exactly k squares in one of the four cardinal directions. A move cannot go beyond the edges of the grid; it does not wrap. What is the minimum number of moves required to get from the top-left corner to the bottom-right corner?

Input

Each input will consist of a single test case. Note that your program may be run multiple times on different inputs. The first line of input contains two space-separated integers n and m (1≤n,m≤500), indicating the size of the grid. It is guaranteed that at least one of n and m is greater than 1. The next n lines will each consist of m digits, with no spaces, indicating the nxm grid. Each digit is between 0 and 9, inclusive. The top-left corner of the grid will be the square corresponding to the first character in the first line of the test case. The bottom-right corner of the grid will be the square corresponding to the last character in the last line of the test case.

Output

Output a single integer on a line by itself representing the minimum number of moves required to get from the top-left corner of the grid to the bottom-right. If it isn’t possible, output -1.

Example

Input:5 4 2120 1203 3113 1120 1110

Output:

6

题意：每次只能走四方方向之一（上下左右），每次走的步数是当前所在位置上的数值，从左上角到右下角求最短需要几步，不能则输出-1。

求最少步数，用bfs，这里注意开一个二维数组vis[i][j]代表从（0，0）到（i，j）经历了几步

#include<iostream>#include<cstring>#include<queue>using namespace std;const int MAXN = 500 + 7;int vis[MAXN][MAXN];//相当于标记，为0我就可以走，此时到达i，j的步数vis【i】【j】就是到达上一个状态的步数加1char gra[MAXN][MAXN];int n, m, cnt;int d[4][2] = {0, -1, 0, 1,1, 0, -1, 0};struct node{    int x, y, val;}next;bool we(int x, int y){    return x >= 0 && x < n && y >= 0 && y < m;}int bfs(int x, int y, int c){    memset(vis, 0, sizeof(vis));    queue<node> q;    node now;    now.x=x;    now.y=y;    now.val=c;    q.push(now);    while(!q.empty())    {        node P = q.front();        q.pop();        for (int i = 0; i < 4; ++i)        {            next.x= P.x + d[i][0]*P.val;            next.y = P.y + d[i][1]*P.val;            next.val=gra[next.x][next.y]-'0';            if (we(next.x,next.y) && !vis[next.x][next.y])            {                vis[next.x][next.y]= (vis[P.x][P.y] + 1);                if(next.x == n-1 && next.y == m-1)                {                    return vis[n-1][m-1];                }                q.push(next);            }        }    }    return -1;}int main(){    while(cin >> n >> m)    {        for(int i = 0; i < n; ++i)        {            cin >> gra[i];        }    cout <<bfs(0, 0, gra[0][0]-'0')<< endl;    }    return 0;}