SPOJ SERGRID Grid BFS搜索算法

A - 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 <bits/stdc++.h> using namespace std; const int MAXN = 500 + 7; int vis[MAXN][MAXN]; char gra[MAXN][MAXN]; int n, m, cnt; int dx[] = {0, -1, 0, 1}; int dy[] = {1, 0, -1, 0}; struct Point { Point(int a, int b, int c) : x(a) , y(b) , val(c) {}; int x, y, val; }; bool isValid(int x, int y) { return x >= 0 && x < n && y >= 0 && y < m; } bool bfs(int x, int y, int c) { memset(vis, 0, sizeof(vis)); queue<Point> q; q.push(Point(x, y, c)); while(!q.empty()) { Point P = q.front(); q.pop(); for (int i = 0; i < 4; ++i) { int _x = P.x + dx[i]*P.val; int _y = P.y + dy[i]*P.val; if (isValid(_x, _y) && !vis[_x][_y]) { Point vertex(_x, _y, gra[_x][_y]-'0'); q.push(vertex); vis[_x][_y] += (vis[P.x][P.y] + 1); if(_x == n-1 && _y == m-1) { return true; } } } } return false; } int main() { ios::sync_with_stdio(false); while(cin >> n >> m) { for(int i = 0; i < n; ++i) { cin >> gra[i]; } if(bfs(0, 0, gra[0][0]-'0')) { cout << vis[n-1][m-1] << endl; } else { cout << -1 << endl; } } return 0; }