ZOJ-1046

来源：互联网发布：极光推送java服务器端编辑：程序博客网时间：2024/06/03 10:21

一般模拟题，不算太麻烦，找出唯一标识字的一像素或两像素，直接枚举硬搞吧

#include<cstdio>#include<vector>using namespace std;namespace{int n, r, c;char s[10][100];bool check_one(vector<vector<pair<int, int> > > &V, size_t index){vector<pair<int, int> > &v = V[index];for (size_t k = 0; k < v.size(); k++){pair<int, int> &p = v[k];int basei = p.first;int basej = p.second - index * (c + 1);bool find = true;for (size_t t = 0; t < V.size(); t++){if (t == index)continue;int i = basei;int j = basej + t * (c + 1);if (s[i][j] != '.'){find = false;break;}}if (find){s[p.first][p.second] = '#';return true;}}return false;}bool check_two(vector<vector<pair<int, int> > > &V, size_t index){vector<pair<int, int> > &v = V[index];for (size_t k = 0; k < v.size(); k++)for (size_t l = k + 1; l < v.size(); l++){pair<int, int> &p1 = v[k];pair<int, int> &p2 = v[l];int firstj = p1.second - index * (c + 1);int secondj = p2.second - index * (c + 1);bool find = true;for (size_t t = 0; t < V.size(); t++){if (t == index)continue;int j1 = firstj + t * (c + 1);int j2 = secondj + t * (c + 1);if (s[p1.first][j1] != '.' && s[p2.first][j2] != '.'){find = false;break;}}if (find){s[p1.first][p1.second] = '#';s[p2.first][p2.second] = '#';return true;}}return false;}}int main(){int cs = 0;vector<vector<pair<int, int> > > V;while (scanf("%d %d %d", &n, &r, &c), n || r || c){printf("Test %d\n", ++cs);getchar();for (int i = 0; i < r; i++)gets(s[i]);V.clear();for (int k = 0; k < n; k++){vector<pair<int, int> > v;int begin = k * (c + 1);for (int i = 0; i < r; i++)for (int j = begin; j < begin + c; j++)if (s[i][j] == 'o')v.push_back(make_pair(i, j));V.push_back(v);}bool res = true;for (size_t k = 0; k < V.size(); k++)if (!check_one(V, k))if (!check_two(V, k)){res = false;break;}if (!res)puts("impossible");elsefor (int i = 0; i < r; i++)puts(s[i]);}}

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