标签:ref printf efi graph har 题解 pac 空白 main
bzoj1661[Usaco2006 Nov]Big Square 巨大正方形
题意:
n*n的图中有一些J点,一些B点和一些空白点,问在空白点添加一个J点所能得到的有4个J点组成最大正方形面积。n≤100。
题解:
枚举两个点,然后根据这两个点组成的边尝试在4个上下两个方向组成四边形。
代码:
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <queue> 5 #define inc(i,j,k) for(int i=j;i<=k;i++) 6 #define maxn 110 7 using namespace std; 8 9 char graph[maxn][maxn]; int n,ans; 10 int main(){ 11 scanf("%d",&n); inc(i,1,n)scanf("%s",graph[i]+1); 12 inc(i,1,n)inc(j,1,n)if(graph[i][j]==‘J‘){ 13 inc(k,i+1,n){ 14 inc(l,1,j){ 15 int cnt=1; 16 if(graph[k][l]==‘B‘)continue; if(graph[k][l]==‘J‘)cnt++; 17 int x=l-j,y=k-i; 18 if(k-x>n||l+y>n||graph[k-x][l+y]==‘B‘)goto jump1; if(graph[k-x][l+y]==‘J‘)cnt++; 19 if(i-x>n||j+y>n||graph[i-x][j+y]==‘B‘)goto jump1; if(graph[i-x][j+y]==‘J‘)cnt++; 20 if(cnt>=3)ans=max(ans,(k-i)*(k-i)+(j-l)*(j-l)); 21 jump1:; 22 if(k+x<1||l-y<1||graph[k+x][l-y]==‘B‘)goto jump2; if(graph[k+x][l-y]==‘J‘)cnt++; 23 if(i+x<1||j-y<1||graph[i+x][j-y]==‘B‘)goto jump2; if(graph[i+x][j-y]==‘J‘)cnt++; 24 if(cnt>=3)ans=max(ans,(k-i)*(k-i)+(j-l)*(j-l)); 25 jump2:; 26 } 27 inc(l,j+1,n){ 28 int cnt=1; 29 if(graph[k][l]==‘B‘)continue; if(graph[k][l]==‘J‘)cnt++; 30 int x=l-j,y=k-i; 31 if(k-x<1||l+y>n||graph[k-x][l+y]==‘B‘)goto jump3; if(graph[k-x][l+y]==‘J‘)cnt++; 32 if(i-x<1||j+y>n||graph[i-x][j+y]==‘B‘)goto jump3; if(graph[i-x][j+y]==‘J‘)cnt++; 33 if(cnt>=3)ans=max(ans,(k-i)*(k-i)+(j-l)*(j-l)); 34 jump3:; 35 if(k+x>n||l-y<1||graph[k+x][l-y]==‘B‘)goto jump4; if(graph[k+x][l-y]==‘J‘)cnt++; 36 if(i+x>n||j-y<1||graph[i+x][j-y]==‘B‘)goto jump4; if(graph[i+x][j-y]==‘J‘)cnt++; 37 if(cnt>=3)ans=max(ans,(k-i)*(k-i)+(j-l)*(j-l)); 38 jump4:; 39 } 40 } 41 } 42 printf("%d",ans); return 0; 43 }
20161023
bzoj1661[Usaco2006 Nov]Big Square 巨大正方形*
标签:ref printf efi graph har 题解 pac 空白 main
原文地址:http://www.cnblogs.com/YuanZiming/p/6013157.html