Codeforces#566 B. Plus from Picture(思维)

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;
}