Ural 1250 Sea Burial 題解_ZenDei技術網路在線

Ural 1250 Sea Burial 題解 [TOC] 題意給定一個$n\times m$的地圖，$.$為水，$\ $為陸，地圖的外部是水(地圖被水包圍)。水為八連通，陸為四聯通。聯通的水稱為海，聯通的陸稱為島。海內可能有島，島內可能有海。給定$x,y$求在包含$(x,y)$(保證$(x,y) ...

Ural 1250 Sea Burial 題解
- 題意
- 輸入
- 題解
- 程式

Ural 1250 Sea Burial 題解

題意

給定一個$n\times m$的地圖，$.$為水，$\#$為陸，地圖的外部是水(地圖被水包圍)。水為八連通，陸為四聯通。聯通的水稱為海，聯通的陸稱為島。海內可能有島，島內可能有海。給定$x,y$求在包含$(x,y)$(保證$(x,y)$為水)的海裡面有多少島。

輸入

第一行包含$m,n,y,x(1\le n,m\le 500,1\le x \le n,1\le y \le m)$

以下若幹行為一個$n\times m$的地圖

題解

考慮BFS或DFS(以下簡稱BFS)

從$(x,y)$BFS，找出包含$(x,y)$的海。
從地圖外部(水)BFS，找出在包含$(x,y)$的海的外面部分。
執行完前兩步，就可以知道包含$(x,y)$的海裡面的部分，數出包含$(x,y)$的海裡面的部分有多少島即可。

Tip: 運用二進位可以使程式簡便。記陸為$1$，島為$2$。設我們需要的值為$x$，當前的值為$y$，只需判斷$(x\&y)$是否大於$0$即可。第1步時$x=2$，第2步時$x=3$(想一想，為什麼，答案最後揭曉)，第3步時$x=1$。

程式

// #pragma GCC optimize(2)
// #pragma G++ optimize(2)
// #pragma comment(linker,"/STACK:102400000,102400000")

// #include <bits/stdc++.h>
#include <map>
#include <set>
#include <list>
#include <array>
#include <cfenv>
#include <cmath>
#include <ctime>
#include <deque>
#include <mutex>
#include <queue>
#include <ratio>
#include <regex>
#include <stack>
#include <tuple>
#include <atomic>
#include <bitset>
#include <cctype>
#include <cerrno>
#include <cfloat>
#include <chrono>
#include <cstdio>
#include <cwchar>
#include <future>
#include <limits>
#include <locale>
#include <memory>
#include <random>
#include <string>
#include <thread>
#include <vector>
#include <cassert>
#include <climits>
#include <clocale>
#include <complex>
#include <csetjmp>
#include <csignal>
#include <cstdarg>
#include <cstddef>
#include <cstdint>
#include <cstdlib>
#include <cstring>
#include <ctgmath>
#include <cwctype>
#include <fstream>
#include <iomanip>
#include <numeric>
#include <sstream>
#include <ccomplex>
#include <cstdbool>
#include <iostream>
#include <typeinfo>
#include <valarray>
#include <algorithm>
#include <cinttypes>
#include <cstdalign>
#include <stdexcept>
#include <typeindex>
#include <functional>
#include <forward_list>
#include <system_error>
#include <unordered_map>
#include <unordered_set>
#include <scoped_allocator>
#include <condition_variable>
// #include <conio.h>
// #include <windows.h>
using namespace std;

typedef long long LL;
typedef unsigned int ui;
typedef unsigned long long ull;
typedef float fl;
typedef double ld;
typedef long double LD;
typedef pair<int,int> pii;
#if (WIN32) || (WIN64) || (__WIN32) || (__WIN64) || (_WIN32) || (_WIN64) || (WINDOWS)
#define lld "%I64d"
#define llu "%I64u"
#else
#define lld "%lld"
#define llu "%llu"
#endif
#define ui(n) ((unsigned int)(n))
#define LL(n) ((long long)(n))
#define ull(n) ((unsigned long long)(n))
#define fl(n) ((float)(n))
#define ld(n) ((double)(n))
#define LD(n) ((long double)(n))
#define char(n) ((char)(n))
#define Bool(n) ((bool)(n))
#define fixpoint(n) fixed<<setprecision(n)

const int INF=1061109567;
const int NINF=-1044266559;
const LL LINF=4557430888798830399;
const ld eps=1e-15;
#define MOD (1000000007)
#define PI (3.1415926535897932384626433832795028841971)

/*
#define MB_LEN_MAX 5
#define SHRT_MIN (-32768)
#define SHRT_MAX 32767
#define USHRT_MAX 0xffffU
#define INT_MIN (-2147483647 - 1)
#define INT_MAX 2147483647
#define UINT_MAX 0xffffffffU
#define LONG_MIN (-2147483647L - 1)
#define LONG_MAX 2147483647L
#define ULONG_MAX 0xffffffffUL
#define LLONG_MAX 9223372036854775807ll
#define LLONG_MIN (-9223372036854775807ll - 1)
#define ULLONG_MAX 0xffffffffffffffffull
*/

#define MP make_pair
#define MT make_tuple
#define All(a) (a).begin(),(a).end()
#define pall(a) (a).rbegin(),(a).rend()
#define log2(x) log(x)/log(2)
#define Log(x,y) log(x)/log(y)
#define SZ(a) ((int)(a).size())
#define rep(i,n) for(int i=0;i<((int)(n));i++)
#define rep1(i,n) for(int i=1;i<=((int)(n));i++)
#define repa(i,a,n) for(int i=((int)(a));i<((int)(n));i++)
#define repa1(i,a,n) for(int i=((int)(a));i<=((int)(n));i++)
#define repd(i,n) for(int i=((int)(n))-1;i>=0;i--)
#define repd1(i,n) for(int i=((int)(n));i>=1;i--)
#define repda(i,n,a) for(int i=((int)(n));i>((int)(a));i--)
#define repda1(i,n,a) for(int i=((int)(n));i>=((int)(a));i--)
#define FOR(i,a,n,step) for(int i=((int)(a));i<((int)(n));i+=((int)(step)))
#define repv(itr,v) for(__typeof((v).begin()) itr=(v).begin();itr!=(v).end();itr++)
#define repV(i,v) for(auto i:v)
#define repE(i,v) for(auto &i:v)
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x) MS(x,0)
#define MINF(x) MS(x,63)
#define MCP(x,y) memcpy(x,y,sizeof(y))
#define sqr(x) ((x)*(x))
#define UN(v) sort(All(v)),v.erase(unique(All(v)),v.end())
#define filein(x) freopen(x,"r",stdin)
#define fileout(x) freopen(x,"w",stdout)
#define fileio(x)\
    freopen(x".in","r",stdin);\
    freopen(x".out","w",stdout)
#define filein2(filename,name) ifstream name(filename,ios::in)
#define fileout2(filename,name) ofstream name(filename,ios::out)
#define file(filename,name) fstream name(filename,ios::in|ios::out)
#define Pause system("pause")
#define Cls system("cls")
#define fs first
#define sc second
#define PC(x) putchar(x)
#define GC(x) x=getchar()
#define Endl PC('\n')
#define SF scanf
#define PF printf

inline int Read()
{
    int X=0,w=0;char ch=0;while(!isdigit(ch)){w|=ch=='-';ch=getchar();}while(isdigit(ch))X=(X<<3)+(X<<1)+(ch^48),ch=getchar();
    return w?-X:X;
}
inline void Write(int x){if(x<0)putchar('-'),x=-x;if(x>9)Write(x/10);putchar(x%10+'0');}

inline LL powmod(LL a,LL b){LL RES=1;a%=MOD;assert(b>=0);for(;b;b>>=1){if(b&1)RES=RES*a%MOD;a=a*a%MOD;}return RES%MOD;}
inline LL gcdll(LL a,LL b){return b?gcdll(b,a%b):a;}
const int dx[]={0,1,0,-1,1,-1,-1,1};
const int dy[]={1,0,-1,0,-1,-1,1,1};
/************************************************************Begin************************************************************/
const int maxn=510;

int n,m,X,Y,s[maxn][maxn],ans; // '#'=>1(01) '.'=>2(10)
bool vis[maxn][maxn];

inline void bfs(int sx,int sy,int status)
{
    vis[sx][sy]=1;
    queue<pair<int,int> > q;q.push({sx,sy});

    while(!q.empty())
    {
        int x=q.front().fs,y=q.front().sc;q.pop();

        rep(i,8)
        {
            if(s[x][y]==1&&i>3) break;

            int cx=x+dx[i],cy=y+dy[i];
            if(cx>0&&cx<=n&&cy>0&&cy<=m&&!vis[cx][cy]&&(s[cx][cy]&status))
            {
                vis[cx][cy]=1;
                q.push({cx,cy});
            }
        }
    }
}

int main()
{
    cin>>m>>n>>Y>>X;
    rep1(i,n) rep1(j,m)
    {
        char c;cin>>c;
        s[i][j]=(c=='#'?1:2);
    }
    
    // step 1
    bfs(X,Y,2);
    
    // step 2
    rep1(i,n)
    {
        bfs(i,0,3);
        bfs(i,m+1,3);
    }
    rep1(j,m)
    {
        bfs(0,j,3);
        bfs(n+1,j,3);
    }
    
    //step 3
    rep1(i,n) rep1(j,m) if(!vis[i][j]&&s[i][j]==1)
    {
        ans++;
        bfs(i,j,1);
    }

    cout<<ans;

    return 0;
}
/*************************************************************End**************************************************************/

DFS(與BFS十分類似)

// #pragma GCC optimize(2)
// #pragma G++ optimize(2)
// #pragma comment(linker,"/STACK:102400000,102400000")

// #include <bits/stdc++.h>
#include <map>
#include <set>
#include <list>
#include <array>
#include <cfenv>
#include <cmath>
#include <ctime>
#include <deque>
#include <mutex>
#include <queue>
#include <ratio>
#include <regex>
#include <stack>
#include <tuple>
#include <atomic>
#include <bitset>
#include <cctype>
#include <cerrno>
#include <cfloat>
#include <chrono>
#include <cstdio>
#include <cwchar>
#include <future>
#include <limits>
#include <locale>
#include <memory>
#include <random>
#include <string>
#include <thread>
#include <vector>
#include <cassert>
#include <climits>
#include <clocale>
#include <complex>
#include <csetjmp>
#include <csignal>
#include <cstdarg>
#include <cstddef>
#include <cstdint>
#include <cstdlib>
#include <cstring>
#include <ctgmath>
#include <cwctype>
#include <fstream>
#include <iomanip>
#include <numeric>
#include <sstream>
#include <ccomplex>
#include <cstdbool>
#include <iostream>
#include <typeinfo>
#include <valarray>
#include <algorithm>
#include <cinttypes>
#include <cstdalign>
#include <stdexcept>
#include <typeindex>
#include <functional>
#include <forward_list>
#include <system_error>
#include <unordered_map>
#include <unordered_set>
#include <scoped_allocator>
#include <condition_variable>
// #include <conio.h>
// #include <windows.h>
using namespace std;

typedef long long LL;
typedef unsigned int ui;
typedef unsigned long long ull;
typedef float fl;
typedef double ld;
typedef long double LD;
typedef pair<int,int> pii;
#if (WIN32) || (WIN64) || (__WIN32) || (__WIN64) || (_WIN32) || (_WIN64) || (WINDOWS)
#define lld "%I64d"
#define llu "%I64u"
#else
#define lld "%lld"
#define llu "%llu"
#endif
#define ui(n) ((unsigned int)(n))
#define LL(n) ((long long)(n))
#define ull(n) ((unsigned long long)(n))
#define fl(n) ((float)(n))
#define ld(n) ((double)(n))
#define LD(n) ((long double)(n))
#define char(n) ((char)(n))
#define Bool(n) ((bool)(n))
#define fixpoint(n) fixed<<setprecision(n)

const int INF=1061109567;
const int NINF=-1044266559;
const LL LINF=4557430888798830399;
const ld eps=1e-15;
#define MOD (1000000007)
#define PI (3.1415926535897932384626433832795028841971)

/*
#define MB_LEN_MAX 5
#define SHRT_MIN (-32768)
#define SHRT_MAX 32767
#define USHRT_MAX 0xffffU
#define INT_MIN (-2147483647 - 1)
#define INT_MAX 2147483647
#define UINT_MAX 0xffffffffU
#define LONG_MIN (-2147483647L - 1)
#define LONG_MAX 2147483647L
#define ULONG_MAX 0xffffffffUL
#define LLONG_MAX 9223372036854775807ll
#define LLONG_MIN (-9223372036854775807ll - 1)
#define ULLONG_MAX 0xffffffffffffffffull
*/

#define MP make_pair
#define MT make_tuple
#define All(a) (a).begin(),(a).end()
#define pall(a) (a).rbegin(),(a).rend()
#define log2(x) log(x)/log(2)
#define Log(x,y) log(x)/log(y)
#define SZ(a) ((int)(a).size())
#define rep(i,n) for(int i=0;i<((int)(n));i++)
#define rep1(i,n) for(int i=1;i<=((int)(n));i++)
#define repa(i,a,n) for(int i=((int)(a));i<((int)(n));i++)
#define repa1(i,a,n) for(int i=((int)(a));i<=((int)(n));i++)
#define repd(i,n) for(int i=((int)(n))-1;i>=0;i--)
#define repd1(i,n) for(int i=((int)(n));i>=1;i--)
#define repda(i,n,a) for(int i=((int)(n));i>((int)(a));i--)
#define repda1(i,n,a) for(int i=((int)(n));i>=((int)(a));i--)
#define FOR(i,a,n,step) for(int i=((int)(a));i<((int)(n));i+=((int)(step)))
#define repv(itr,v) for(__typeof((v).begin()) itr=(v).begin();itr!=(v).end();itr++)
#define repV(i,v) for(auto i:v)
#define repE(i,v) for(auto &i:v)
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x) MS(x,0)
#define MINF(x) MS(x,63)
#define MCP(x,y) memcpy(x,y,sizeof(y))
#define sqr(x) ((x)*(x))
#define UN(v) sort(All(v)),v.erase(unique(All(v)),v.end())
#define filein(x) freopen(x,"r",stdin)
#define fileout(x) freopen(x,"w",stdout)
#define fileio(x)\
    freopen(x".in","r",stdin);\
    freopen(x".out","w",stdout)
#define filein2(filename,name) ifstream name(filename,ios::in)
#define fileout2(filename,name) ofstream name(filename,ios::out)
#define file(filename,name) fstream name(filename,ios::in|ios::out)
#define Pause system("pause")
#define Cls system("cls")
#define fs first
#define sc second
#define PC(x) putchar(x)
#define GC(x) x=getchar()
#define Endl PC('\n')
#define SF scanf
#define PF printf

inline int Read()
{
    int X=0,w=0;char ch=0;while(!isdigit(ch)){w|=ch=='-';ch=getchar();}while(isdigit(ch))X=(X<<3)+(X<<1)+(ch^48),ch=getchar();
    return w?-X:X;
}
inline void Write(int x){if(x<0)putchar('-'),x=-x;if(x>9)Write(x/10);putchar(x%10+'0');}

inline LL powmod(LL a,LL b){LL RES=1;a%=MOD;assert(b>=0);for(;b;b>>=1){if(b&1)RES=RES*a%MOD;a=a*a%MOD;}return RES%MOD;}
inline LL gcdll(LL a,LL b){return b?gcdll(b,a%b):a;}
const int dx[]={0,1,0,-1,1,-1,-1,1};
const int dy[]={1,0,-1,0,-1,-1,1,1};
/************************************************************Begin************************************************************/
const int maxn=510;

int n,m,X,Y,s[maxn][maxn],ans; // '#'=>1(01) '.'=>2(10)
bool vis[maxn][maxn];

inline void dfs(int x,int y,int status)
{
    vis[x][y]=1;

    rep(i,8)
    {
        if(s[x][y]==1&&i>3) break;

        int cx=x+dx[i],cy=y+dy[i];
        if(cx>0&&cx<=n&&cy>0&&cy<=m&&!vis[cx][cy]&&(s[cx][cy]&status)) dfs(cx,cy,status);
    }
}

int main()
{
    cin>>m>>n>>Y>>X;
    rep1(i,n) rep1(j,m)
    {
        char c;cin>>c;
        s[i][j]=(c=='#'?1:2);
    }
    
    // step 1
    dfs(X,Y,2);
    
    // step 2
    rep1(i,n)
    {
        dfs(i,0,3);
        dfs(i,m+1,3);
    }
    rep1(j,m)
    {
        dfs(0,j,3);
        dfs(n+1,j,3);
    }
    
    //step 3
    rep1(i,n) rep1(j,m) if(!vis[i][j]&&s[i][j]==1)
    {
        ans++;
        dfs(i,j,1);
    }

    cout<<ans;

    return 0;
}
/*************************************************************End**************************************************************/

Tip's answer: 第2步是需要找出在包含$(x,y)$的海的外面部分，而外面部分不分海陸，$x=3$即$x=(11)_2$，這樣$1\&3$與$2\&3$都大於$0$了。