dfs | 洛谷 | P1101

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https://www.luogu.org/problemnew/show/P1101

1. 用二重循环构造一个八连通增量
2. for外夹紧的dfs

#include <bits/stdc++.h>
#define FF(a,b) for(int a=0;a<b;a++)
#define F(a,b) for(int a=1;a<=b;a++)
#define LEN 110
#define bug(x) cout<<#x<<"="<<x<<endl;

using namespace std;

typedef struct point
{
    int x,y;
    point(int x=0,int y=0):x(x),y(y){}
}point;
point pt[LEN];
char word[LEN][LEN];
char stand[]="yizhong";
int vis[LEN][LEN];//保存路径
int dir[8][2];
int n;

void build_dir()
{
    int k=0;
    FF(i,3)FF(j,3){
        if(i==1 && j==1) continue;
        dir[k][0]=i-1;
        dir[k][1]=j-1;
        k++;
    }
}

bool legal(int x){
    if(x>=0 && x<n)
        return true;
    else return false;
}

void dfs(int x,int y,int k,int cur){
    if(cur==7){
        FF(i,7){
            vis[pt[i].x][pt[i].y]=1;
        }
        return;
    }
    int dx=dir[k][0];
    int dy=dir[k][1];
    if(legal(x) && legal(y) && word[x][y]==stand[cur]){
        pt[cur]=point(x,y);
        dfs(x+dx,y+dy,k,cur+1);
    }
}


int main()
{
    build_dir();
    freopen("./in","r",stdin);

    scanf("%d",&n);
    for(int i=0; i<n; i++)
        scanf("%s",word[i]);
    FF(i,n)FF(j,n)if(word[i][j]=='y'){
        FF(k,8){
            int tx=i+dir[k][0];
            int ty=j+dir[k][1];
            if(word[tx][ty]=='i'){
                dfs(i,j,k,0);
            }
        }
    }
    FF(i,n){
        FF(j,n) putchar(vis[i][j]?word[i][j]:'*' );
        puts("");
    }

    return 0;
}