cf-edu#5-C The Labyrinth-dfs记忆化搜搜

http://codeforces.com/contest/616/problem/C

 给出一个n*m图，

如 ：

*.*.*.*.*

求 每个*周围连在一起的点“.”共有多少个，个数+上自己本身，也就是+1。

窝是直接dfs记忆化搜索，求出每个点“.”的联通量，然后标记在同一个联通块中的点通通为mark(唯一标记)

输出的时候，判断*周围四个方向的空点”.“所含的联通量，并判断一下不要重复计算了 【在同一联通块的点】

int n,m;char tm[1005][1005]; int num[1005][1005];int dx[8]={0,0,1,-1};int dy[8]={1,-1,0,0};int vis[1005][1005];queue< int* > sb;int who[1005][1005]; int dfs(int i,int j,int mmark){if (num[i][j]) return num[i][j];num[i][j]=1; vis[i][j]=1; who[i][j]=mmark;int *t=&num[i][j];sb.push(t);int k,x,y;for (k=0;k<4;k++){x=i+dx[k];y=j+dy[k]; if (x>=1&&x<=n&&y>=1&&y<=m&&tm[x][y]=='.'&&vis[x][y]==0){num[i][j]+=dfs(x,y,mmark);} }return num[i][j]; }int main(){cin>>n>>m;int i,j;int minnn=2147483647;for (i=1;i<=n;i++){scanf("%s",tm[i]+1);} int x,y,k;int mark=1;for (i=1;i<=n;i++){for (j=1;j<=m;j++){if (tm[i][j]=='*') continue;num[i][j]=dfs(i,j,mark++);int *tt;while(!sb.empty()){tt=sb.front();sb.pop();*tt=num[i][j];}}} for (i=1;i<=n;i++){for (j=1;j<=m;j++){if (tm[i][j]=='.') printf(".");else{int tmp=1;int aa[5];int tmp_ok=0;for (k=0;k<4;k++){x=i+dx[k];y=j+dy[k]; if (x>=1&&x<=n&&y>=1&&y<=m){if (tm[x][y]=='.' ){int flag=0;for (int h=1;h<=tmp_ok;h++){if (who[x][y]==aa[h]) flag=1;}if (flag)continue;tmp+=num[x][y];aa[++tmp_ok]=who[x][y];}} }printf("%d",tmp%10);}}printf("\n");} return 0;}

后来写的代码

：
#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <algorithm>#include <iostream>#include <queue>#include <map>#include <set>#include <vector>using namespace std; const int N=200005;int id=0;int dx[4]={0,0,1,-1};int dy[4]={1,-1,0,0,};char tm[1005][1005];int col[1005][1005];int num[1005*1005];     int n,m;int ans[1005][1005];void dfs(int x,int y,int tmp_id){if (col[x][y]) return ;if (tmp_id==0) col[x][y]=++id,tmp_id=id;elsecol[x][y]=tmp_id; num[tmp_id]++;for (int i=0;i<4;i++){int tx=x+dx[i];int ty=y+dy[i];if (!(tx>=1&&tx<=n&&ty>=1&&ty<=m)) continue;if (tm[tx][ty]=='*')continue;dfs(tx,ty,id);}}int main(){int i,j,k,tx,ty;cin>>n>>m;for (i=1;i<=n;i++){scanf("%s",tm[i]+1);}for (i=1;i<=n;i++){for (j=1;j<=m;j++){if (tm[i][j]=='.')dfs(i,j,0);}}set<int >sb;for (i=1;i<=n;i++){for (j=1;j<=m;j++){int sum=0;if (tm[i][j]=='.')continue;sb.clear();for (k=0;k<4;k++){int tx=i+dx[k];int ty=j+dy[k];if (!(tx>=1&&tx<=n&&ty>=1&&ty<=m)) continue;if (tm[tx][ty]=='.'&&sb.find(col[tx][ty])==sb.end()) {sum+=num[col[tx][ty]];sb.insert(col[tx][ty]);}}ans[i][j]=sum+1;}}for (i=1;i<=n;i++){for (j=1;j<=m;j++){if (ans[i][j])printf("%d",ans[i][j]%10);else printf(".");}printf("\n");}return 0;}