hdu 5301 Buildings 2015 Multi-University Training Contest 2
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最受不了这种智商题了~~o(>_<)o ~~又是一开始就想错了><
如果没有(x,y),那么最优解肯定是width/2,如果有了黑点,就必须要从width/2多出格子来补偏离了水平中轴线的部分。
case 1:竖直方向的添格子填补,此时带来的ans1=max(x-1,n-x)
X ↑ ↑ ↑ ↑case 2:水平方向的添格子填补,此时带来的ans2=min(y,m-y+1)。因为方案→和方案←选择一个即可,剩下的照例用竖直方向width/2的划分方法。
X →→ or ←←←←最后选择较小的一个case,因为其他部分是用width/2填补的,两者最大值为最终答案。
另外边为奇数的正方形且(x,y)位于正中心的case要特判,画个图可以发现不会存在(width+1)/2的划分。
#include<iostream>#include<stdio.h>#include<cstdio>#include<stdlib.h>#include<vector>#include<string>#include<cstring>#include<cmath>#include<algorithm>#include<stack>#include<queue>#include<ctype.h>#include<map>#include<time.h>#include<bitset>#include<set>#include<list>using namespace std;//hdu 5301const int maxn=100010;int n;int m;int x;int y;int main(){ freopen("input.txt","r",stdin); while(scanf("%d %d %d %d",&n,&m,&x,&y)!=EOF) { memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); if(n>m) { swap(n,m); swap(x,y); } if(n==m&&n%2!=0&&m%2!=0) { if(x==(n+1)/2&&y==(m+1)/2) { printf("%d\n",x-1); continue; } } if(m<=2||n<=2) { printf("1\n"); continue; } int ans1=max(x-1,n-x); int ans2=min(y,m-y+1); int ans3=(n%2==0?n/2:n/2+1); int ans=max(ans3,min(ans1,ans2)); printf("%d\n",ans); } return 0;}
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