【BZOJ 2252】 矩阵距离

【题目链接】

https://www.lydsy.com/JudgeOnline/problem.php?id=2252

【算法】

将所有是”1“的点入队，然后广度优先搜索，即可

【代码】

#include<bits/stdc++.h>
using namespace std;
#define MAXN 1010

const int dx[] = {,,-,};
const int dy[] = {-,,,};

int i,j,n,m,tx,ty;
int dist[MAXN][MAXN];
char s[MAXN][MAXN];
char val;
pair<int,int> cur;
queue< pair<int,int> > q;

inline bool ok(int x,int y)
{
        return x >=  && x <= n && y >=  && y <= m;
}

int main()
{
        memset(dist,,sizeof(dist));
        scanf("%d%d",&n,&m);
        for (i = ; i <= n; i++) scanf("%s",s[i]+);
        for (i = ; i <= n; i++)
        {
                for (j = ; j <= m; j++)
                {
                        if (s[i][j] == '')
                        {
                                q.push(make_pair(i,j));
                                dist[i][j] = ;
                        }
                }
        }
        while (!q.empty())
        {
                cur = q.front();
                q.pop();
                for (i = ; i < ; i++)
                {
                        tx = cur.first + dx[i];
                        ty = cur.second + dy[i];
                        if (ok(tx,ty) && dist[tx][ty] == -)
                        {
                                dist[tx][ty] = dist[cur.first][cur.second] + ;
                                q.push(make_pair(tx,ty));
                        }
                }
        }
        for (i = ; i <= n; i++)
        {
                for (j = ; j <= m; j++)
                {
                        printf("%d ",dist[i][j]);
                }
                printf("\n");
        }

        return ;

}