标签:view -- sed back div nbsp efi cin pac
把每一列中相邻的 . 缩为一个点 作为二分图的左边
把每一行中相邻的 . 缩为一个点 作为二分图的右边
然后求最大匹配即可
这题用匈牙利足够了。。。然而。。我用了hk。。。有点大材小用的感觉///
#include <iostream> #include <cstdio> #include <cstring> #include <queue> #include <vector> #include <cmath> #include <algorithm> #define mem(a, b) memset(a, b, sizeof(a)) using namespace std; const int maxn = 10010, INF = 0x7fffffff; int dx[maxn], dy[maxn], cx[maxn], cy[maxn], used[maxn]; int row[5][5], col[5][5]; int nx, ny, dis; vector<int> G[40005]; char str[5][5]; int n; int bfs() { queue<int> Q; dis = INF; mem(dx, -1); mem(dy, -1); for(int i=1; i<=nx; i++) { if(cx[i] == -1) { Q.push(i); dx[i] = 0; } } while(!Q.empty()) { int u = Q.front(); Q.pop(); if(dx[u] > dis) break; for(int v=0; v<G[u].size(); v++) { int i = G[u][v]; if(dy[i] == -1) { dy[i] = dx[u] + 1; if(cy[i] == -1) dis = dy[i]; else { dx[cy[i]] = dy[i] + 1; Q.push(cy[i]); } } } } return dis != INF; } int dfs(int u) { for(int v=0; v<G[u].size(); v++) { int i = G[u][v]; if(!used[i] && dy[i] == dx[u] + 1) { used[i] = 1; if(cy[i] != -1 && dis == dy[i]) continue; if(cy[i] == -1 || dfs(cy[i])) { cy[i] = u; cx[u] = i; return 1; } } } return 0; } int hk() { int res = 0; mem(cx, -1); mem(cy, -1); while(bfs()) { mem(used, 0); for(int i=1; i<=nx; i++) { if(cx[i] == -1 && dfs(i)) res++; } } return res; } int main() { while(cin>> n && n) { for(int i=0; i<100; i++) G[i].clear(); mem(row, -1); mem(col, -1); nx = ny = 1; for(int i=0; i<n; i++) { cin>> str[i]; } for(int i=0; i<n; i++) { for(int j=0; j<n; j++) { if(str[i][j] == ‘.‘ && row[i][j] == -1) { for(int k=j; str[i][k]==‘.‘ && k<n; k++) row[i][k] = nx; nx++; } if(str[j][i] == ‘.‘ && col[j][i] == -1) { for(int k=j; str[k][i]==‘.‘ && k<n; k++) col[k][i] = ny; ny++; } } } nx -= 1, ny -= 1; for(int i=0; i<n; i++) for(int j=0; j<n; j++) if(str[i][j] == ‘.‘) G[row[i][j]].push_back(nx + col[i][j]), G[nx + col[i][j]].push_back(row[i][j]); cout<< hk() <<endl; } return 0; }
搜索写法:
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #define mem(a,b) memset(a,b,sizeof(a)) using namespace std; const int maxn = 6, INF = 0xfffffff; typedef long long LL; char str[maxn][maxn]; int vis[maxn][maxn]; int n, minn; int check(int x,int y) { for(int i=x-1; i>=0; --i) { if(vis[i][y]) return 0; if(str[i][y] == ‘X‘) break; } for(int i=y-1; i>=0; --i) { if(vis[x][i]) return 0; if(str[x][i] == ‘X‘) break; } return 1; } void dfs(int inx, int k) { if(inx == n*n) { minn = max(k, minn); return; } int x = inx / n; int y = inx % n; if(str[x][y] == ‘.‘ && check(x,y)) { vis[x][y] = 1; dfs(inx+1, k+1); vis[x][y] = 0; } dfs(inx+1, k); } int main() { while(cin>>n && n) { minn = -INF; mem(vis,0); mem(str,0); for(int i=0;i<n;i++) cin>>str[i]; dfs(0,0); cout<<minn<<endl; } return 0; }
标签:view -- sed back div nbsp efi cin pac
原文地址:https://www.cnblogs.com/WTSRUVF/p/9309488.html