BZOJ 1047 理想的正方形(单调队列)

刚开始用二维RMQ直接给超内存了。。。

用单调队列可以做到O(n^2)的复杂度。具体是先把每行用单调队列处理一下。再把处理后的用列单调队列处理下。

 

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-9
# define MOD 12345678
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int res=0, flag=0;
    char ch;
    if((ch=getchar())=='-') flag=1;
    else if(ch>='0'&&ch<='9') res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')  res=res*10+(ch-'0');
    return flag?-res:res;
}
void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=1005;
//Code begin...

int val[N][N], row_max[N][N], row_min[N][N], col_max[N][N], col_min[N][N], que[N], head, tail;

int main ()
{
    int n, m, a;
    scanf("%d%d%d",&n,&m,&a);
    FOR(i,1,n) FOR(j,1,m) scanf("%d",&val[i][j]);
    FOR(i,1,n) {
        head=-1; tail=0;
        FOR(j,1,m) {
            while (head>=tail&&val[i][j]>=val[i][que[head]]) --head;
            que[++head]=j;
            if (que[head]-que[tail]>=a) ++tail;
            if (j>=a) row_max[i][j-a+1]=val[i][que[tail]];
        }
        head=-1; tail=0;
        FOR(j,1,m) {
            while (head>=tail&&val[i][j]<=val[i][que[head]]) --head;
            que[++head]=j;
            if (que[head]-que[tail]>=a) ++tail;
            if (j>=a) row_min[i][j-a+1]=val[i][que[tail]];
        }
    }
    FOR(j,1,m-a+1) {
        head=-1; tail=0;
        FOR(i,1,n) {
            while (head>=tail&&row_max[i][j]>=row_max[que[head]][j]) --head;
            que[++head]=i;
            if (que[head]-que[tail]>=a) ++tail;
            if (i>=a) col_max[i-a+1][j]=row_max[que[tail]][j];
        }
        head=-1; tail=0;
        FOR(i,1,n) {
            while (head>=tail&&row_min[i][j]<=row_min[que[head]][j]) --head;
            que[++head]=i;
            if (que[head]-que[tail]>=a) ++tail;
            if (i>=a) col_min[i-a+1][j]=row_min[que[tail]][j];
        }
    }
    int ans=INF;
    FOR(i,1,n-a+1) FOR(j,1,m-a+1) ans=min(ans,col_max[i][j]-col_min[i][j]);
    printf("%d\n",ans);
    return 0;
}