BZOJ 1047: [HAOI2007]理想的正方形

题目

单调队列是个很神奇的东西,我以前在博客写过(吧)

我很佩服rank里那些排前几的大神,700ms做了时限10s的题,简直不能忍。(但是我还是不会写

我大概一年半没写单调队列,也有可能根本没有写过

放两个程序,第一个TLE,第二个3s+直接通过

哦对了!单调队列的复杂度为O(n),很经典的证明呢

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int MAXab=1000+7,MAXn=100+7,inf=2147483647;
typedef int MATRIX[MAXn][MAXab][MAXab];
MATRIX fmin,fmax;
inline int max(int a,int b){return a>b?a:b;}
inline int min(int a,int b){return a>b?b:a;}
inline int min4(int a,int b,int c,int d){return min(min(min(a,b),c),d);}
inline int max4(int a,int b,int c,int d){return max(max(max(a,b),c),d);}

int main(){
	//freopen("b1047.in","r",stdin);
	int a,b,n;
	scanf("%d%d%d",&a,&b,&n);
	for (int i=1;i<=a;i++)
		for (int j=1;j<=b;j++) {
			scanf("%d",&fmin[1][i][j]);
			fmax[1][i][j]=fmin[1][i][j];
		}
	int ans=inf;
	for (int k=2;k<=n;k++){         //kは决策です!
		for (int i=1;i<=a-k+1;i++)
			for (int j=1;j<=b-k+1;j++){
				fmin[k][i][j]=
					min4(fmin[k-1][i][j],fmin[k-1][i+1][j],
					fmin[k-1][i][j+1],fmin[k-1][i+1][j+1]);
				fmax[k][i][j]=
					max4(fmax[k-1][i][j],fmax[k-1][i+1][j],
					fmax[k-1][i][j+1],fmax[k-1][i+1][j+1]);
				if (k==n) ans=min(fmax[k][i][j]-fmin[k][i][j],ans);
			}
	}
	printf("%d",ans);
	return 0;
}
/*优先队列!*/
#include <iostream>
#include <cstdio>
using namespace std;
const int MAXn=100+7,MAXab=1000+7;
const int inf=0x7fffffff;
int n,a,b,fmin[MAXab][MAXab],fmax[MAXab][MAXab];

inline int max(int a,int b){return a>b?a:b;}
inline int min(int a,int b){return a<b?a:b;}

int main(){
//	freopen("b1047.in","r",stdin);
//	freopen("b1047.out","w",stdout);
	scanf("%d%d%d",&a,&b,&n);
	for (int i=1;i<=a;i++){
		int ts,ws,tb,wb,qs[MAXab]={0},qb[MAXab]={-inf},v[MAXab]={0};
		ts=ws=tb=wb=1;
		for (int j=1;j<=b;j++){
			scanf("%d",&v[j]);
			while (ws-->ts&&v[j]<v[qs[ws]]);ws++;qs[ws++]=j;
			if (qs[ws-1]-qs[ts]+1>n) ts++;
			while (wb-->tb&&v[j]>v[qb[wb]]);wb++;qb[wb++]=j;
			if (qb[wb-1]-qb[tb]+1>n) tb++;
			if (j>=n) fmin[i][j-n+1]=v[qs[ts]],fmax[i][j-n+1]=v[qb[tb]];
		}
	}

	int ans=inf;
	for (int i=1;i<=a-n+1;i++){
		for (int j=1;j<=b-n+1;j++){
			int minans=inf,maxans=-inf;
			for (int k=0;k<n;k++){
				minans=min(minans,fmin[i+k][j]);
				maxans=max(maxans,fmax[i+k][j]);
			}
			ans=min(ans,maxans-minans);
		}
	}
	cout<<ans<<endl;
	return 0;
}

posted on 2016-07-26 07:45  Chuckqgz  阅读(...)  评论(...编辑  收藏

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