题目链接
解题思路
简单贪心，由于每个格子始终不超过 \(9\) 个史莱姆，因此对于每四个格子 \(a_{i-1,j},a_{i+1,j},a_{i,j-1},a_{i,j+1}\)，我们只需要减去 \(\min(a_{i-1,j},a_{i+1,j},a_{i,j-1},a_{i,j+1})\) 即可，容易证明这样贪心是正确的。
参考代码
#include<bits/stdc++.h>
using namespace std;
#define map unordered_map
#define forl(i,a,b) for(register long long i=a;i<=b;i++)
#define forr(i,a,b) for(register long long i=a;i>=b;i--)
#define lc(x) x<<1
#define rc(x) x<<1|1
#define cin(x) scanf("%lld",&x)
#define cout(x) printf("%lld",x)
#define lowbit(x) x&-x
#define pb push_back
#define pf push_front
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
#define endl '\n'
#define QwQ return 0;
#define ll long long
ll n,m,a[510][510],minn;
int main()
{
//IOS;
cin>>n>>m;
forl(i,1,n)
forl(j,1,m)
scanf("%1lld",&a[i][j]);
forl(i,1,n)
{
forl(j,1,m)
minn=min({a[i-1][j],a[i+1][j],a[i][j-1],a[i][j+1]}),cout<<minn,a[i-1][j]-=minn,a[i+1][j]-=minn,a[i][j-1]-=minn,a[i][j+1]-=minn;
cout<<endl;
}
/******************/
/*while(L<q[i].l) */
/* del(a[L++]);*/
/*while(L>q[i].l) */
/* add(a[--L]);*/
/*while(R<q[i].r) */
/* add(a[++R]);*/
/*while(R>q[i].r) */
/* del(a[R--]);*/
/******************/
QwQ;
}