题目大意：给你地图，让你判断需要多少水才可以将农场灌满。
题解：显然用并查集比较容易，将可以连通的并起来，最后输出连通块的数目即可，一开始我用字母分类讨论发现很麻烦，于是参考别人的博客发现，直接自己写一个矩阵，然后处理一下读入数据会比较简单：
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#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; } |
原文：http://www.cnblogs.com/forever97/p/3549352.html