題目大意：給你地圖，讓你判斷需要多少水才可以將農場灌滿。

題解：顯然用并查集比較容易，將可以連通的并起來，最后輸出連通塊的數目即可，一開始我用字母分類討論發現很麻煩，于是參考別人的博客發現，直接自己寫一個矩陣，然后處理一下讀入數據會比較簡單：

#include <cstring> #include <cstdio> #include <iostream> using namespace std; int R[11][11]={{0,0,0,0,0,0,0,0,0,0,0}, {1,0,1,0,0,1,1,1,1,0,1},{0,0,0,0,0,0,0,0,0,0,0}, {1,0,1,0,0,1,1,1,1,0,1},{0,0,0,0,0,0,0,0,0,0,0}, {1,0,1,0,0,1,1,1,1,0,1},{1,0,1,0,0,1,1,1,1,0,1}, {0,0,0,0,0,0,0,0,0,0,0},{1,0,1,0,0,1,1,1,1,0,1}, {1,0,1,0,0,1,1,1,1,0,1},{1,0,1,0,0,1,1,1,1,0,1}}; int U[11][11]={{0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0},{1,1,0,0,1,0,1,1,0,1,1}, {1,1,0,0,1,0,1,1,0,1,1},{1,1,0,0,1,0,1,1,0,1,1}, {0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0}, {1,1,0,0,1,0,1,1,0,1,1},{1,1,0,0,1,0,1,1,0,1,1}, {1,1,0,0,1,0,1,1,0,1,1},{1,1,0,0,1,0,1,1,0,1,1}}; string map[55]; int f[3000]; void init(int n) { int i; for(i=0;i<=n;i++) f[i]=i; } int sf(int i) { int j=i; while(j!=f[j]) { j=f[j]; } return f[i]=j; } int Union(int x,int y) { x=sf(x); y=sf(y); if(x==y) return 0; else { f[x]=y; return 1; } } int main() { int n,m; while(scanf("%d%d",&m,&n),m!=-1&&n!=-1) { int i,j; init(n*m); for(i=0;i<m;i++) cin>>map[i]; for(i=0;i<m;i++) for(j=1;j<n;j++) if(R[map[i][j-1]-'A'][map[i][j]-'A']) Union(i*n+j-1,i*n+j); for(i=0;i<n;i++) for(j=1;j<m;j++) if(U[map[j-1][i]-'A'][map[j][i]-'A']) Union((j-1)*n+i,j*n+i); int count=0; for(i=0;i<n*m;i++) { if(f[i]==i) count++; } printf("%d\n",count); } return 0; }

轉載于:https://www.cnblogs.com/forever97/p/3549352.html

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