题意很简明吧？
枚举的矩形下边界和右端点即右下角，来确定矩形位置；
每一个纵列开一个单调队列，记录从 i-n+1 行到 i 行每列的最大值和最小值，矩形下边界向下推移的时候维护一下；
然后在记录的每一列的最大值和最小值上，跑滑动窗口，即用单调队列维护区间 [ j-n+1 , j ] 的最大值和最小值；
相当于维护了一个矩形的最大值和最小值
#include<cstdio> #include<iostream> #include<queue> #define R register int using namespace std; inline int g() {R ret=0,fix=1; register char ch; while(!isdigit(ch=getchar())) fix=ch=='-'?-1:fix;do ret=ret*10+(ch^48); while(isdigit(ch=getchar())); return ret*fix; } inline int abs(int x) {return x>0?x:-x;} int a,b,n,ans=0x3f3f3f3f; int vl[1010][1010]; deque<int> q[1010],qq[1010],p,pp; signed main() {a=g(),b=g(),n=g();for(R i=1;i<=a;++i) for(R j=1;j<=b;++j) vl[i][j]=g();for(R i=1;i<=a;++i) {for(R j=1;j<=b;++j) {while(q[j].size()&&vl[q[j].back()][j]<vl[i][j]) q[j].pop_back();while(q[j].size()&&q[j].front()+n<=i) q[j].pop_front();q[j].push_back(i);//cout<<i<<"hang"<<j<<"lie"<<vl[i][q[j].front()]<<" ";//cout<<"mx"<<vl[q[j].front()][j]<<" "; } for(R j=1;j<=b;++j) {while(qq[j].size()&&vl[qq[j].back()][j]>vl[i][j]) qq[j].pop_back();while(qq[j].size()&&qq[j].front()+n<=i) qq[j].pop_front();qq[j].push_back(i);//cout<<"mn"<<vl[qq[j].front()][j]<<" ";}//cout<<'\n';if(i<n) continue; p.clear();pp.clear();for(R j=1;j<=b;++j) {while(p.size()&&vl[q[p.back()].front()][p.back()]<vl[q[j].front()][j]) p.pop_back();while(p.size()&&p.front()+n<=j) p.pop_front();p.push_back(j);while(pp.size()&&vl[qq[pp.back()].front()][pp.back()]>vl[qq[j].front()][j]) pp.pop_back();while(pp.size()&&pp.front()+n<=j) pp.pop_front();pp.push_back(j);if(j<n) continue;ans=min(abs(vl[q[p.front()].front()][p.front()]-vl[qq[pp.front()].front()][pp.front()]),ans);}}printf("%d\n",ans); }
2019.04.06
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