思路:限制当前能剪下的最大格子数,保证能得到最少数目。IDA*的典型运用。
AC代码
#include <cstdio> #include <cmath> #include <algorithm> #include <cstring> #include <utility> #include <string> #include <iostream> #include <map> #include <set> #include <vector> #include <queue> #include <stack> using namespace std; #pragma comment(linker, "/STACK:1024000000,1024000000") #define eps 1e-10 #define inf 0x3f3f3f3f #define PI pair<int, int> typedef long long LL; const int maxn = 10 + 5; int vis[maxn][maxn], a[maxn][maxn]; int n, m, maxd, goal; const int dx[] = {0,0,-1,1}; const int dy[] = {1,-1,0,0}; // IDA* int w; int dfs(int x, int y, int sum, int cnt) { if(sum == goal) return 1; if(cnt >= maxd) { w = min(w, sum); return 0; } vis[x][y] = 1; for(int i = 0; i < 4; ++i) { int px = x + dx[i], py = y + dy[i]; if(px < 0 || py < 0 || px >= n || py >= m || vis[px][py]) continue; if(sum + a[px][py] > goal) continue; if(dfs(px, py, sum + a[px][py], cnt + 1)) return 1; } vis[x][y] = 0; return 0; } int main() { while(scanf("%d%d", &m, &n) == 2) { int sum = 0; for(int i = 0; i < n; ++i) for(int j = 0; j < m; ++j) { scanf("%d", &a[i][j]); sum += a[i][j]; } if(sum & 1) { printf("0\n"); continue; } memset(vis, 0, sizeof(vis)); goal = sum / 2; int flag = 0; for(maxd = 1; maxd < n*m; ++maxd) { w = inf; if(dfs(0, 0, a[0][0], 1)) { flag = 1; printf("%d\n", maxd); break; } if(w > goal) break; } if(!flag) printf("0\n"); } return 0; }
如有不当之处欢迎指出!