UVa572 Oil Deposits DFS求连通块

 

技巧:遍历8个方向

for(int dr = -1; dr <= 1; dr++)
    for(int dc = -1; dc <= 1; dc++)
      if(dr != 0 || dc != 0) dfs(r+dr, c+dc, id);

我的解法:

#include<cstdio>
#include<cstring>
#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
const int N=102;
char buf[N][N];
int m, n;
int cnt;
int dr[]={0, 0, 1, 1, 1, -1, -1, -1};
int dc[]={1, -1, -1, 0, 1, -1, 0, 1};

void dfs(int r, int c)
{
    buf[r][c]='*';
    for(int i=0;i<8;i++)
    {
        if(buf[r+dr[i]][c+dc[i]]=='@')
            dfs(r+dr[i], c+dc[i]);
    }
    
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("./uva572.in", "r", stdin);
#endif
    while(cin>>m>>n && m)
    {
        memset(buf, '*', sizeof buf);
        cnt=0;
        for(int i=1;i<=m;i++)
        {
            cin>>(buf[i]+1);
        }

        for(int i=1;i<=m;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(buf[i][j]=='@')
                {
                    dfs(i, j);
                    cnt++;
                }
            }
        }
        cout<<cnt<<endl;
    }

    return 0;
}

 

lrj的解法在统计的时候把连通块的序号也求出来了,保存在idx中。

// UVa572 Oil Deposits
// Rujia Liu
// 题意:输入一个字符矩阵,统计字符@组成多少个四连块
#include<cstdio>
#include<cstring>
const int maxn = 100 + 5;

char pic[maxn][maxn];
int m, n, idx[maxn][maxn];

void dfs(int r, int c, int id) {
  if(r < 0 || r >= m || c < 0 || c >= n) return;
  if(idx[r][c] > 0 || pic[r][c] != '@') return;
  idx[r][c] = id;
  for(int dr = -1; dr <= 1; dr++)
    for(int dc = -1; dc <= 1; dc++)
      if(dr != 0 || dc != 0) dfs(r+dr, c+dc, id);
}

int main() {
  while(scanf("%d%d", &m, &n) == 2 && m && n) {
    for(int i = 0; i < m; i++) scanf("%s", pic[i]);
    memset(idx, 0, sizeof(idx));
    int cnt = 0;
    for(int i = 0; i < m; i++)
      for(int j = 0; j < n; j++)
        if(idx[i][j] == 0 && pic[i][j] == '@') dfs(i, j, ++cnt);
    printf("%d\n", cnt);
  }
  return 0;
}

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