链接：https://www.nowcoder.com/acm/contest/134/C
来源：牛客网
第一行两个整数n，m。n表示方格的规格，m表示最初病菌所在的格子数。(1 ≤ n ≤ 1000, 0 < m < n)。
如果最终所有的方格都会被感染，输出 YES。
否则输出 NO。
NO
分析：考虑每个病菌只会影响到周围上下左右四个点，且只能在他上下左右四个斜方向上有病菌的情况下，但是这些被感染的点还会继续影响下面的，所以直接dfs找出所有可能被感染的点就可以了
AC代码：
#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <bitset>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define ls (r<<1)
#define rs (r<<1|1)
#define debug(a) cout << #a << " " << a << endl
using namespace std;
typedef long long ll;
const ll maxn = 1e3+10;
const double eps = 1e-8;
const ll mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
ll mapn[maxn][maxn];
ll n, m;
void dfs( ll x, ll y ) {
if( x-1 > 0 && y-1 > 0 && mapn[x-1][y-1] ) {
if( !mapn[x][y-1] ) {
mapn[x][y-1] = 1;
dfs(x,y-1);
}
if( !mapn[x-1][y] ) {
mapn[x-1][y] = 1;
dfs(x-1,y);
}
}
if( x+1 <= n && y-1 > 0 && mapn[x+1][y-1] ) {
if( !mapn[x][y-1] ) {
mapn[x][y-1] = 1;
dfs(x,y-1);
}
if( !mapn[x+1][y] ) {
mapn[x+1][y] = 1;
dfs(x+1,y);
}
}
if( x+1 <= n && y+1 <= n && mapn[x+1][y+1] ) {
if( !mapn[x+1][y] ) {
mapn[x+1][y] = 1;
dfs(x+1,y);
}
if( !mapn[x][y+1] ) {
mapn[x][y+1] = 1;
dfs(x,y+1);
}
}
if( x-1 > 0 && y+1 <= n && mapn[x-1][y+1] ) {
if( !mapn[x][y+1] ) {
mapn[x][y+1] = 1;
dfs(x,y+1);
}
if( !mapn[x-1][y] ) {
mapn[x-1][y] = 1;
dfs(x-1,y);
}
}
}
int main() {
ios::sync_with_stdio(0);
cin >> n >> m;
memset(mapn,0,sizeof(mapn));
vector<pair<ll,ll> > e;
while( m -- ) {
ll x, y;
cin >> x >> y;
mapn[x][y] = 1;
e.push_back(make_pair(x,y));
}
for( ll i = 0; i < e.size(); i ++ ) {
dfs(e[i].first,e[i].second);
}
bool flag = true;
for( ll i = 1; i <= n; i ++ ) {
for( ll j = 1; j <= n; j ++ ) {
if( !mapn[i][j] ) {
flag = false;
break;
}
}
}
if( flag ) {
cout << "YES" << endl;
} else {
cout << "NO" << endl;
}
return 0;
}
原文：https://www.cnblogs.com/l609929321/p/9531922.html