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UVA - 572 Oil Deposits(dfs)

时间:2017-01-09 18:06:34      阅读:132      评论:0      收藏:0      [点我收藏+]

题意:求连通块个数。

分析:dfs。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
typedef long long ll;
typedef unsigned long long llu;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const double pi = acos(-1.0);
const double eps = 1e-8;
const int MAXN = 100 + 10;
const int MAXT = 40 + 10;
using namespace std;
int cnt;
int m, n;
int vis[MAXN][MAXN];
char a[MAXN][MAXN];
int judge(int i, int j){
    return i >= 0 && i < m && j >= 0 && j < n;
}
void dfs(int x, int y){
    vis[x][y] = 1;
    for(int i = 0; i < 8; ++i){
        int tmpx = x + dr[i];
        int tmpy = y + dc[i];
        if(judge(tmpx, tmpy) && !vis[tmpx][tmpy] && a[tmpx][tmpy] == @){
            dfs(tmpx, tmpy);
        }
    }
}
int main(){
    while(scanf("%d%d", &m, &n) == 2){
        if(!m && !n) return 0;
        memset(vis, 0, sizeof vis);
        cnt = 0;
        for(int i = 0; i < m; ++i)
            scanf("%s", a[i]);
        for(int i = 0; i < m; ++i){
            for(int j = 0; j < n; ++j){
                if(a[i][j] == @ && !vis[i][j]){
                    ++cnt;
                    dfs(i, j);
                }
            }
        }
        printf("%d\n", cnt);
    }
}

 

UVA - 572 Oil Deposits(dfs)

原文:http://www.cnblogs.com/tyty-Somnuspoppy/p/6265844.html

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