POJ 1164
这道题的题目叙述着实唬人,感觉太多无关信息了
不过说来惭愧,偷了个小懒去看了下别人怎么翻译的,这部分小痛苦直接跳过了
简单DFS应用
#include <iostream>
#include <algorithm>
#include <queue>
#include <string>
#include <vector>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <stack>
#include <map>
#include <set>
using namespace std;
const int maxn= 55;
int m, n;
int cas[maxn][maxn];
bool vis[maxn][maxn];
int step[4][2]= {{0, -1}, {-1, 0}, {0, 1}, {1, 0}};
int DFS(int x, int y)
{
if (x< 1 || x> m || y< 1 || y>n){
return 0;
}
int ret= 1;
for (int i= 0; i< 4; ++i){
if (cas[x][y] & (1<<i)){
continue;
}
int nx= x+step[i][0], ny= y+step[i][1];
if (!vis[nx][ny]){
vis[nx][ny]= 1;
ret+= DFS(x+step[i][0], y+step[i][1]);
}
}
return ret;
}
int main(int argc, char const *argv[])
{
scanf("%d %d", &m, &n);
for (int i= 1; i<= m; ++i){
for (int j= 1; j<= n; ++j){
scanf("%d", cas[i]+j);
}
}
int ans=0, mx= 0;
memset(vis, 0, sizeof(vis));
for (int i= 1; i<= m; ++i){
for (int j= 1; j<= n; ++j){
if (!vis[i][j]){
++ans;
vis[i][j]= 1;
mx= max(mx, DFS(i, j));
}
}
}
printf("%d\n%d\n", ans, mx);
return 0;
}