USACO 2.1 The Castle-漫水填充

题目描述

In a stroke of luck almost beyond imagination, Farmer John was sent a ticket to the Irish Sweepstakes (really a lottery) for his birthday. This ticket turned out to have only the winning number for the lottery! Farmer John won a fabulous castle in the Irish countryside.

Bragging rights being what they are in Wisconsin, Farmer John wished to tell his cows all about the castle. He wanted to know how many rooms it has and how big the largest room was. In fact, he wants to take out a single wall to make an even bigger room.

Your task is to help Farmer John know the exact room count and sizes.

The castle floorplan is divided into M (wide) by N (1 <=M,N<=50) square modules. Each such module can have between zero and four walls. Castles always have walls on their "outer edges" to keep out the wind and rain.

Consider this annotated floorplan of a castle:

     1   2   3   4   5   6   7
   #############################
 1 #   |   #   |   #   |   |   #
   #####---#####---#---#####---#   
 2 #   #   |   #   #   #   #   #
   #---#####---#####---#####---#
 3 #   |   |   #   #   #   #   #   
   #---#########---#####---#---#
 4 # ->#   |   |   |   |   #   #   
   ############################# 

#  = Wall     -,|  = No wall
-> = Points to the wall to remove to
     make the largest possible new room

By way of example, this castle sits on a 7 x 4 base. A "room" includes any set of connected "squares" in the floor plan. This floorplan contains five rooms (whose sizes are 9, 7, 3, 1, and 8 in no particular order).

Removing the wall marked by the arrow merges a pair of rooms to make the largest possible room that can be made by removing a single wall.

The castle always has at least two rooms and always has a wall that can be removed.

Solution

容易发现墙的方向1,2,4,8都是2的幂次方,我们可以用位运算来判断每个格子的哪个方向有墙,我们用一个三维数组来记录每个格子中有墙的方向,\( a[i][j][k] \)代表(i,j)这个格子的\(2^k\)方向是否有墙。

先考虑第一、二问,我们使用一个dfs漫水填充(i,j)所在的联通块,在calc函数中枚举所有的格子,统计出联通块的个数以及最大联通块的面积

然后考虑第三、四问,我们继续暴力枚举所有的格子以及其各个方向,如果有墙,则拆除这面墙并重新统计

Code

/*
PROB: castle
LANG: C++
ID: yizhuor1
*/
#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair

using namespace std;

typedef long long LL;
typedef pair<int, int> pii;

template <typename T> inline int Chkmax (T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline int Chkmin (T &a, T b) { return a > b ? a = b, 1 : 0; }

template <typename T> inline T read(){
    T sum = 0;
    int fl = 1,ch = getchar();
    for(; !isdigit(ch); ch = getchar()) if(ch == '-') fl = -1;
    for(; isdigit(ch); ch = getchar()) sum = (sum << 3) + (sum << 1) + ch - '0';
    return sum * fl;
}

const int maxn = 50 + 5;

int n,m;
int sum,cnt,ansx,ansy,ansfx,maxans;
int a[maxn][maxn][5],vis[maxn][maxn];
int dx[] = {0,-1,0,1};
int dy[] = {-1,0,1,0};

void dfs(int x,int y){  //就是个简单的漫水填充
    vis[x][y] = 1;
    cnt++;
    for(int i = 0; i < 4; i++){
        int nx = x + dx[i];
        int ny = y + dy[i];
        if(a[x][y][i]) continue;
        if(nx >= 1 && nx <= n && ny >= 1 && ny <= m && !vis[nx][ny]){
            dfs(nx,ny);
        }
    }
}

void calc(int x,int y,int fx){  //拆除(x,y)的fx方向的墙
    memset(vis,0,sizeof(vis));
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= m; j++){
            if(vis[i][j]) continue;
            cnt = 0;
            dfs(i,j);
            sum++;
            if(cnt > maxans){
                maxans = cnt;
                ansx = x;
                ansy = y;
                ansfx = fx;
            }
        }
    }
}

inline void Solve (){
    calc(0,0,0);
    printf("%d\n%d\n",sum,maxans);
    maxans = 0;
    for(int j = 1; j <= m; j++){
        for(int i = n; i >= 1; i--){
            for(int k = 1; k < 3; k++){
                if(!a[i][j][k]) continue;
                a[i][j][k] = 0;
                if(k == 1) a[i - 1][j][3] = 0;
                if(k == 2) a[i][j + 1][0] = 0;
                calc(i,j,k);
                a[i][j][k] = 1;
                if(k == 1) a[i - 1][j][3] = 1;
                if(k == 2) a[i][j + 1][0] = 1;
            }
        }
    }
    printf("%d\n",maxans);
    char ch;
    if(ansfx == 1){
        ch = 'N';
    }else ch = 'E';
    printf("%d %d %c\n",ansx,ansy,ch);
}

inline void Input (){
    m = read<int>();
    n = read<int>();
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= m; j++){
            int x;
            x = read<int>();
            if(x & 1) a[i][j][0] = 1;
            if(x & 2) a[i][j][1] = 1;
            if(x & 4) a[i][j][2] = 1;
            if(x & 8) a[i][j][3] = 1;
        }
    }
}

int main(){
    Input();
    Solve();
    return 0;
}

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