Codeforces Round #501 E2. Stars Drawing (Hard Edition) 暴力枚举+标记

题意：

定义一种“星”，问有多少个，输出坐标和大小，，不合法时输出-1

思路：

他的星可以重叠，所以我们枚举每个星的中心点，尽量往外找，找到这个点为中心的最大大小的星

存起来最后 对整个矩阵标记，检查合法情况

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1000 + 7;
int n, m;
char s[maxn][maxn];
vector<pair<int,pair<int,int> > > ans;
bool ok(int x, int y) {
    if(x > 0 && x <= n && y > 0 && y <= m && s[x][y] == '*')
        return true;
    else
        return false;
}
void solve(int x, int y) {
    int x1,y1,x2,y2,x3,y3,x4,y4,anss = 0;
    for(int i = 1; ; ++i) {
        x1 = x+i, y1 = y;
        x2 = x-i, y2 = y;
        x3 = x,   y3 = y+i;
        x4 = x,   y4 = y-i;
        if(ok(x1,y1) && ok(x2,y2) && ok(x3,y3) && ok(x4,y4)) {
            anss = i;
        }
        else break;
    }
    if(anss) {
        pair<int,int> p1 = make_pair(x,y);
        pair<int,pair<int,int> > p2 = make_pair(anss,p1);
        ans.push_back(p2);
    }

}
int main() {
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; ++i) {
        scanf("%s", s[i]+1);
    }
    for(int i = 1; i <= n; ++i) {
        for(int j = 1; j <= m; ++j) {
            if(s[i][j] == '*') {
                solve(i,j);
            }
        }
    }
    for(auto i : ans) {
        int x = i.second.first, y = i.second.second, d = i.first;
        int x1,y1,x2,y2,x3,y3,x4,y4;
        s[x][y] = '.';
        for(int j = 1; j <= d; ++j) {
            x1 = x+j, y1 = y;
            x2 = x-j, y2 = y;
            x3 = x,   y3 = y+j;
            x4 = x,   y4 = y-j;
            s[x1][y1] = '.';
            s[x2][y2] = '.';
            s[x3][y3] = '.';
            s[x4][y4] = '.';
        }
    }
    bool f = true;
    for(int i = 1; i <= n; ++i) {
        for(int j = 1; j <= m; ++j) {
            if(s[i][j] == '*') {
                printf("-1\n"); return 0;
                //f = false; break;
            }
        }
    }
    printf("%d\n", ans.size());
    for(auto i : ans) {
        printf("%d %d %d\n", i.second.first, i.second.second, i.first);
    }
    return 0;
}