CF 542C - n^4枚举

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题目链接:http://codeforces.com/contest/1130/problem/C

解题思路:

一个数组维护(r1,c1)能到达的点，一个数组维护(r2,c2)能到达的点，数组两两枚举取最小即可。

#include<algorithm>
#include<cstdio>
#include <iostream>
#include <map>
using namespace std;
typedef long long ll;
const int mx = 1e2 + 10;
const int mod = 1e9 + 7;
const int fx[4] = {1,0,0,-1};
const int fy[4] = {0,1,-1,0};
int n,r1,c2,r2,c1;
char str[mx][mx];
bool a[mx][mx],b[mx][mx];
void dfs(int x,int y,bool p[][mx])
{
	if(p[x][y]) return ;
	p[x][y] = 1;
	for(int i=0;i<4;i++){
		int dx = x + fx[i];
		int dy = y + fy[i];
		if(dx<=0||dy<=0||dx>n||dy>n) continue;
		if(str[dx][dy]=='0') dfs(dx,dy,p);
	}
}
int Get(int x,int y)
{
	int ans = 1e9;
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			if(b[i][j])
			ans = min(ans,(x-i)*(x-i) + (y-j)*(y-j));
		}
	}
	return ans;
}
int main()
{
	scanf("%d",&n);
	scanf("%d%d%d%d",&r1,&c1,&r2,&c2);
	for(int i=1;i<=n;i++) scanf("%s",str[i]+1);
	dfs(r1,c1,a);
	dfs(r2,c2,b);
	int ans = 1e9;
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			if(a[i][j])
			ans = min(ans,Get(i,j));
		}
	}
	printf("%d\n",ans);
    return 0;
}