[USACOFeb2021S]ComfortableCows

Farmer Nhoj's pasture can be regarded as a large 2D grid of square "cells" (picture a huge chessboard). Initially, the pasture is empty.
Farmer Nhoj will add N (1≤N≤10⁵) cows to the pasture one by one. The ith cow will occupy a cell (x_i,y_i) that is distinct from the cells occupied by all other cows (0≤x_i,y_i≤1000).

A cow is said to be "comfortable" if it is horizontally or vertically adjacent to exactly three other cows. Unfortunately, cows that are too comfortable tend to lag in their milk production, so Farmer Nhoj wants to add additional cows until no cow (including the ones that he adds) is comfortable. Note that the added cows do not necessarily need to have x and y coordinates in the range 0…1000.

For each i in the range 1…N, please output the minimum number of cows Farmer Nhoj would need to add until no cows are comfortable if initially, the pasture started with only cows 1…i.

For i=4, Farmer Nhoj must add an additional cow at (2,1) to make the cow at (1,1) uncomfortable.

For i=9, the best Farmer Nhoj can do is place additional cows at (2,0), (3,0), (2,−1), and (2,3).

Problem credits: Benjamin Qi

#include <bits/stdc++.h> using namespace std; int dx[5]={0,1,0,-1,0}; int dy[5]={0,0,1,0,-1}; int m[3005][3005]; int jsq; int n; void dfs(int x,int y) { int flg=0; int ta1,ta2; if(!m[x][y]) return ; for(int i=1;i<=4;i++) { int xx=x+dx[i]; int yy=y+dy[i]; // if(xx<0||xx>n||yy<n||yy>n) // continue; if(m[xx][yy]) flg++; else ta1=xx,ta2=yy; } if(flg!=3) return ; else { m[ta1][ta2]=1; jsq++; for(int i=0;i<=4;i++) dfs(ta1+dx[i],ta2+dy[i]); } } int main() { cin>>n; while(n--) { int x,y; cin>>x; cin>>y; x+=1000; y+=1000; if(m[x][y]) jsq--; m[x][y]=1; for(int i=0;i<=4;i++) { int xx=x+dx[i]; int yy=y+dy[i]; dfs(xx,yy); } cout<<jsq<<endl; } return 0; }

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