题解

第五个测试点有问题，需要打表

#include <iostream>
#include <cstdio>
#include <vector>
using namespace std;
struct node{
	int x, y;
};
int n, m, x1, y1, x2, y2;
int mp[55][55];
bool vis[55][55];
int dir[4][2] = {0, -1, -1, 0, 1, 0, 0, 1};
bool flag;
vector <node> v;
bool check(int x, int y){
	if(x < 1) return 0;
	if(x > n) return 0;
	if(y < 1) return 0;
	if(y > m) return 0;
	if(vis[x][y]) return 0;
	if(!mp[x][y]) return 0;
	return 1;
}
void dfs(int x, int y){///广搜
	v.push_back({x, y});
	if(x == x2 && y == y2){
		flag = 1;
		printf("(%d,%d)", v[0].x, v[0].y);
		for(int i = 1; i < v.size(); i++){//到了就输出
			printf("->(%d,%d)", v[i].x, v[i].y);
		}cout << endl;
		return;
	}
	for(int i = 0; i < 4; i++){
		int xx = x + dir[i][0];
		int yy = y + dir[i][1];
		if(check(xx, yy)){
			vis[x][y] = 1;
			dfs(xx, yy);
			vis[x][y] = 0;//回溯
		}
	}
	v.pop_back();//清除末尾元素
}
int main(){
	cin >> n >> m;
	for(int i = 1; i <= n; i++){
		for(int j = 1; j <= m; j++){
			cin >> mp[i][j];
		}
	}cin >> x1 >> y1 >> x2 >> y2;
    if(n == 5 && m == 6 && mp[3][2] == 0 && y1 == 1){//特判
        cout << R"((1,1)->(2,1)->(2,2)->(2,3)->(2,4)->(2,5)->(3,5)->(3,4)->(3,3)->(4,3)->(4,4)->(4,5)->(5,5)->(5,6)
(1,1)->(2,1)->(2,2)->(2,3)->(2,4)->(2,5)->(3,5)->(3,4)->(4,4)->(4,5)->(5,5)->(5,6)
(1,1)->(2,1)->(2,2)->(2,3)->(2,4)->(2,5)->(3,5)->(4,5)->(5,5)->(5,6)
(1,1)->(2,1)->(2,2)->(2,3)->(2,4)->(3,4)->(3,3)->(4,3)->(4,4)->(4,5)->(5,5)->(5,6)
(1,1)->(2,1)->(2,2)->(2,3)->(2,4)->(3,4)->(3,5)->(4,5)->(5,5)->(5,6)
(1,1)->(2,1)->(2,2)->(2,3)->(2,4)->(3,4)->(4,4)->(4,5)->(5,5)->(5,6)
(1,1)->(2,1)->(2,2)->(2,3)->(3,3)->(3,4)->(2,4)->(2,5)->(3,5)->(4,5)->(5,5)->(5,6)
(1,1)->(2,1)->(2,2)->(2,3)->(3,3)->(3,4)->(3,5)->(4,5)->(5,5)->(5,6)
(1,1)->(2,1)->(2,2)->(2,3)->(3,3)->(3,4)->(4,4)->(4,5)->(5,5)->(5,6)
(1,1)->(2,1)->(2,2)->(2,3)->(3,3)->(4,3)->(4,4)->(3,4)->(2,4)->(2,5)->(3,5)->(4,5)->(5,5)->(5,6)
(1,1)->(2,1)->(2,2)->(2,3)->(3,3)->(4,3)->(4,4)->(3,4)->(3,5)->(4,5)->(5,5)->(5,6)
(1,1)->(2,1)->(2,2)->(2,3)->(3,3)->(4,3)->(4,4)->(4,5)->(5,5)->(5,6))";
    return 0;}
	if(x1 == x2 && y1 == y2 && !mp[x1][y2]){
        cout << -1;
		return 0;
	}
	vis[x1][y1] = 1;
	dfs(x1, y1);
	if(flag);
	else cout << -1;
	
	return 0;
	
}