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  • luogu2216 [HAOI2007]理想的正方形

    先对于每一行中长度为 n 的列用单调队列搞出它们的最小/大值,再将这些长度为 n 的列想象成点再对行跑一遍

    #include <iostream>
    #include <cstring>
    #include <cstdio>
    using namespace std;
    int a, b, n, r[1005][1005], qwq[1005], qaq[1005], hwq, twq, haq, taq;
    //qwq xiao ,qaq da
    int hzxz[1005][1005], hzdz[1005][1005];
    //hzxz[i][j]: the minist value in n cols which starts from j in row i
    int zzxz[1005][1005], zzdz[1005][1005];
    int ans=0x3f3f3f3f, t;
    int main(){
    	cin>>a>>b>>n;
    	for(int i=1; i<=a; i++)
    		for(int j=1; j<=b; j++)
    			scanf("%d", &r[i][j]);
    	for(int i=1; i<=a; i++){
    		hwq = haq = 1;
    		twq = taq = 0;
    		for(int j=1; j<=b; j++){
    			t = max(1, j-n+1);
    			while(hwq<=twq && qwq[hwq]<=j-n)	hwq++;
    			while(haq<=taq && qaq[haq]<=j-n)	haq++;
    			while(hwq<=twq && r[i][qwq[twq]]>r[i][j])	twq--;
    			while(haq<=taq && r[i][qaq[taq]]<r[i][j])	taq--;
    			qwq[++twq] = j;
    			qaq[++taq] = j;
    			hzxz[i][t] = r[i][qwq[hwq]];
    			hzdz[i][t] = r[i][qaq[haq]];
    		}
    	}
    	for(int i=1; i<=b-n+1; i++){
    		hwq = haq = 1;
    		twq = taq = 0;
    		for(int j=1; j<=a; j++){
    			t = max(1, j-n+1);
    			while(hwq<=twq && qwq[hwq]<=j-n)	hwq++;
    			while(haq<=taq && qaq[haq]<=j-n)	haq++;
    			while(hwq<=twq && hzxz[qwq[twq]][i]>hzxz[j][i])	twq--;
    			while(haq<=taq && hzdz[qaq[taq]][i]<hzdz[j][i])	taq--;
    			qwq[++twq] = j;
    			qaq[++taq] = j;
    			zzxz[t][i] = hzxz[qwq[hwq]][i];
    			zzdz[t][i] = hzdz[qaq[haq]][i];
    		}
    	}
    	for(int i=1; i<=a-n+1; i++)
    		for(int j=1; j<=b-n+1; j++)
    			ans = min(ans, zzdz[i][j]-zzxz[i][j]);
    	cout<<ans<<endl;
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/poorpool/p/8303696.html
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