先预处理出来所有点所对应的上下左右各有多少*,再用两个差分数组,一个横向一个纵向,来还原最多可以填满的网格
#include<bits/stdc++.h> using namespace std; char graph[1010][1010]; int up[1010][1010]; int dn[1010][1010]; int ri[1010][1010]; int le[1010][1010]; int judgex[1010][1010]; int judgey[1010][1010]; int judge[1010][1010]; struct ans{ int x,y,l; }; vector<ans> vec; int main() { int n,m; scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) { getchar(); for(int j=1;j<=m;j++) { scanf("%c",&graph[i][j]); } } for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { if(graph[i][j]=='*') { up[i][j]=up[i-1][j]+1; le[i][j]=le[i][j-1]+1; } } } for(int i=n;i>=1;i--) { for(int j=m;j>=1;j--) { if(graph[i][j]=='*') { dn[i][j]=dn[i+1][j]+1; ri[i][j]=ri[i][j+1]+1; } } } for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { if(graph[i][j]=='*') { int index=min(min(up[i][j]-1,dn[i][j]-1),min(ri[i][j]-1,le[i][j]-1)); if(index>0) { ans now; now.x=i; now.y=j; now.l=index; vec.push_back(now); judgex[i-index][j]++; judgex[i+index+1][j]--; judgey[i][j-index]++; judgey[i][j+index+1]--; } } } } for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { judgex[i][j]+=judgex[i-1][j]; judgey[i][j]+=judgey[i][j-1]; } } for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { judge[i][j]=judgex[i][j]+judgey[i][j]; } } for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { if(graph[i][j]=='*'&&judge[i][j]==0) { printf("-1\n"); return 0; } } } printf("%d\n",vec.size()); for(int i=0;i<vec.size();i++) { printf("%d %d %d\n",vec[i].x,vec[i].y,vec[i].l); } }