USACO 2.1.1The Castle

// the castle.cpp : 定义控制台应用程序的入口点。///*ID: maiyuet1PROG: castleLANG: C++*///#include "stdafx.h"#include <iostream>#include <cstdio>#include <cstring>#include <queue>using namespace std;const int dir[4][2] = {0,-1,-1,0,0,1,1,0}; //北 西 东 南const int MAXN = 55;int num_house;typedef struct POINT{int x;int y;int wall[4];int cnt;int sign;}Point;Point  p[MAXN][MAXN];int visited[MAXN][MAXN];int row,column;int bfs(int x, int y){int cnt = 1;queue<Point>Q;queue<Point>Save;Point start;start.x = x;start.y = y;Q.push(start);Save.push(start);while(!Q.empty()){Point flag = Q.front();Q.pop();for(int i=0; i<4; i++){if(!p[flag.x][flag.y].wall[i]) //没有墙{Point temp;temp.x = flag.x + dir[i][0];temp.y = flag.y + dir[i][1];if(temp.x >= 1 && temp.x <= row && temp.y >= 1 && temp.y <= column && !visited[temp.x][temp.y]){cnt++;visited[temp.x][temp.y] = 1;Q.push(temp);Save.push(temp);}}}}while(!Save.empty()){Point top = Save.front();Save.pop();p[top.x][top.y].cnt = cnt;p[top.x][top.y].sign = num_house;}return cnt;}int main(){int a;int max_house;int ans_x,ans_y;char ans_d;freopen("castle.in","r",stdin);freopen("castle.out","w",stdout);while(cin>>column>>row){num_house = max_house = 0;memset(visited,0,sizeof(visited));for(int i=1; i<=row; i++){for(int j=1; j<=column; j++){cin>>a;p[i][j].wall[0] = a & 1; p[i][j].wall[1] = a & 2;p[i][j].wall[2] = a & 4;p[i][j].wall[3] = a & 8;}}for(int i=1; i<=row; i++){for(int j=1; j<=column; j++){if(!visited[i][j]){visited[i][j] = 1;num_house++;int num = bfs(i,j);if(num > max_house){max_house = num;}}}}int ans = 0;for(int j=1; j<=column; j++) //从左下角开始尝试,因为有多解的时候选择最靠西，再最靠南{for(int i=row; i>=1; i--){for(int k=1; k<=2; k++)  //只往东和北的方向推墙{if(p[i][j].wall[k])  //有墙{int a = i + dir[k][0];int b = j + dir[k][1];if(a >= 1 && a <= row && b >= 1 && b <= column){if(p[i][j].sign != p[a][b].sign && p[i][j].cnt + p[a][b].cnt > ans){//cout<<i<<" "<<j<<" "<<p[i][j].cnt<<" "<<a<<" "<<b<<" "<<p[a][b].cnt<<endl;ans = p[i][j].cnt + p[a][b].cnt;ans_x = i;ans_y = j;if(k == 1){ans_d = 'N';}else{ans_d = 'E';}}}}}}}cout<<num_house<<endl<<max_house<<endl<<ans<<endl<<ans_x<<" "<<ans_y<<" "<<ans_d<<endl;}return 0;}