刚看到一个做法很棒，枚举十字中点然后向四周扩展“十字”,并计数，如果数量等于*总数就输出YES,否则输出NO.
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<map>
#include<queue>
#include<vector>
#include<string>
#include<fstream>
using namespace std;
#define rep(i, a, n) for(int i = a; i <= n; ++ i);
#define per(i, a, n) for(int i = n; i >= a; -- i);
typedef long long ll;
const int N = 600;
const int mod = 998244353;
const double Pi = acos(- 1.0);
const ll INF = 1e9;
const int G = 3, Gi = 332748118;
ll qpow(ll a, ll b) { ll res = 1; while(b){ if(b & 1) res = (res * a) % mod; a = (a * a) % mod; b >>= 1;} return res; }
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
//
int n, m;
char s[N];
int a[N][N];
int main()
{
int all = 0;
scanf("%d%d",&n,&m);
for(int i = 1; i <= n; ++ i){
scanf("%s",s + 1);
for(int j = 1; j <= m; ++ j){
a[i][j] = (s[j] == '*') ? 1 : 0;
if(a[i][j]) all ++;
}
}
int flag = 0, t;
for(int i = 1; i <= n; ++ i){
for(int j = 1; j <= m; ++ j){
if(!a[i][j] || !a[i + 1][j] || !a[i - 1][j] || !a[i][j + 1] || !a[i][j - 1]) continue;
int num = 1;
t = 1; while(i + t <= n && a[i + t][j]) t ++, num ++;
t = -1; while(i + t >= 1 && a[i + t][j]) t --, num ++;
t = 1; while(j + t <= m && a[i][j + t]) t ++, num ++;
t = -1; while(j + t >= 1 && a[i][j + t]) t --, num ++;
if(num == all){
flag = 1;
break;
}
}
if(flag) break;
}
if(flag) printf("YES
");
else printf("NO
");
return 0;
}
暴力代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<map>
#include<queue>
#include<vector>
#include<string>
#include<fstream>
using namespace std;
#define rep(i, a, n) for(int i = a; i <= n; ++ i);
#define per(i, a, n) for(int i = n; i >= a; -- i);
typedef long long ll;
const int N = 600;
const int mod = 998244353;
const double Pi = acos(- 1.0);
const ll INF = 1e9;
const int G = 3, Gi = 332748118;
ll qpow(ll a, ll b) { ll res = 1; while(b){ if(b & 1) res = (res * a) % mod; a = (a * a) % mod; b >>= 1;} return res; }
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
//
int n, m;
char s[N];
int a[N][N];
int dp[N][N][5];
int vis[N][N];
bool check(int x, int y, int flag, int to){
if(x < 1 || x > n || y < 1 || y > m || !a[x][y]) return false;
int val = 0;
if(!flag){
for(int i = 0; i < 4; ++ i){
val += (dp[x][y][i] > 1) ? 1 : 0;
}
if(val == 4 && a[x][y]) return true;
}
vis[x][y] = 1;
bool tpp = false;
if(to == 0){
if(a[x - 1][y] && !vis[x - 1][y]){
if(check(x - 1, y, 0, to)) tpp = true;
}
if(a[x + 1][y] && !vis[x + 1][y]){
if(check(x + 1, y, 0, to)) tpp = true;
}
}
else{
if(a[x][y - 1] && !vis[x][y - 1]){
if(check(x, y - 1, 0, to)) tpp = true;
}
if(a[x][y + 1] && !vis[x][y + 1]){
if(check(x, y + 1, 0, to)) tpp = true;
}
}
return tpp;
}
void solve(){
for(int i = 1; i <= n; ++ i){
for(int j = 1; j <= m; ++ j){
if(!a[i][j]) continue;
dp[i][j][0] = dp[i - 1][j][0] + 1;
dp[i][j][1] = dp[i][j - 1][1] + 1;
}
}
for(int i = n; i >= 1; -- i){
for(int j = m; j >= 1; -- j){
if(!a[i][j]) continue;
dp[i][j][2] = dp[i + 1][j][2] + 1;
dp[i][j][3] = dp[i][j + 1][3] + 1;
}
}
int num = 0, flag = 0;
for(int i = 1; i <= n; ++ i){
for(int j = 1; j <= m; ++ j){
if(!a[i][j]) continue;
int tp = 0;
if(dp[i][j][0] > 1) tp ++;
if(dp[i][j][1] > 1) tp ++;
if(dp[i][j][2] > 1) tp ++;
if(dp[i][j][3] > 1) tp ++;
if(tp == 4) num ++;
else if(tp == 0 || tp == 3){
flag = 1;
break;
}
else if(tp == 2){
if((a[i - 1][j] != a[i + 1][j]) || (a[i][j - 1] != a[i][j + 1])){
flag = 1;
break;
}
memset(vis, 0, sizeof(vis));
if(a[i - 1][j]) check(i, j, 1, 0);
else check(i, j, 1, 1);
}
else{
memset(vis, 0, sizeof(vis));
int to;
if(a[i - 1][j] || a[i + 1][j]) to = 0;
else to = 1;
if(!check(i, j, 1, to)) {
flag = 1;
break;
}
}
}
if(flag) break;
}
if(!flag && num == 1) printf("YES
");
else printf("NO
");
}
int main()
{
scanf("%d%d",&n,&m);
for(int i = 1; i <= n; ++ i){
scanf("%s",s + 1);
for(int j = 1; j <= m; ++ j){
a[i][j] = (s[j] == '*') ? 1 : 0;
}
}
solve();
return 0;
}