codeforces----193A Cutting Figure
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道广说这个题可以用最小割点集来做,但是点太多,肯定得超时。一开始有点儿没想明白,第二天早晨起来仔细一考虑,稍微分析一下就能得出,只有三种情况,一种是‘#’数量少于3,这种情况不可分割,第二种情况是只需要割一个点就可以的,最后一种情况是只需要割两个点就可以,所以果断上来判断‘#’个数,如果小于3,直接输出-1,否则枚举割一个点的位置,用dfs搜一下看‘#’被分成几部分,发现大于1的话直接返回,输出1,再则直接输出2.
#include <iostream>#include <cstring>#define N 55using namespace std;int n, m;int map[N][N], vis[N][N];void dfs(int x, int y){vis[x][y] = 1;if(0 <= x - 1 && map[x - 1][y] && !vis[x - 1][y])dfs(x - 1, y);if(x + 1 < n && map[x + 1][y] && !vis[x + 1][y])dfs(x + 1, y);if(0 <= y - 1 && map[x][y - 1] && !vis[x][y - 1])dfs(x, y - 1);if(y + 1 < m && map[x][y + 1] && !vis[x][y + 1])dfs(x, y + 1);}int work(int x, int y){int i, j,cnt=0;memset(vis, 0, sizeof(vis));vis[x][y] = 1;for(i = 0; i < n; i ++)for(j = 0; j < m; j ++)if(map[i][j] && !vis[i][j]){dfs(i, j);cnt ++;}return cnt;}int solve(){int i, j;for(i = 0; i < n; i ++)for(j = 0; j < m; j ++){if(work(i, j) > 1)return 1;}return 0;}int main(){int i, j, cnt = 0;char buf;cin >> n >> m;for(i = 0; i < n; i ++)for(j = 0; j < m; j ++){cin >> buf;map[i][j] = buf == '#';cnt += map[i][j];}if(cnt < 3)cout << -1 << endl;else if(solve())cout << 1 << endl;elsecout << 2 << endl;return 0;}
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