UVALive
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题目链接:https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2868
用暴力也过了:
#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<queue>#include<vector>#include<map>#include<string>#include<set>#define LL long long#define MAX(a,b) (a>b?a:b)#define MIN(a,b) (a<b?a:b)using namespace std;int a[1050][1050];int n,m;int ans = 0;int test(int x, int y, int len){ for(int i = x; i<x+len; i++) for(int j = y; j<y+len; j++) { if(a[i][j]==0) return 0; } return 1;}int main(){ while(scanf("%d%d",&n,&m) && (m||n)) { ans = 0; for(int i = 1; i<=n; i++) for(int j = 1; j<=m; j++) scanf("%d",&a[i][j]); for(int i = 1; i<=n; i++) for(int j = 1; j<=m; j++) { if(a[i][j]) for(int len = 1; j+len-1<=m && i+len-1<=n; len++) { if(len<=ans)//这一步很重要,如果长度都比ans小,那再去比较也无意义了。少了这一句就少时了。 continue; if(test(i,j,len)) ans = MAX(ans,len); else break; } } printf("%d\n",ans); } return 0;}
然后使用dp,画图找找规律。dp[i][j]为当前格子作为右下角能达到的最大正方形边长,可知dp[i][j] 只与dp[i-1][j] ,dp[i][j-1] ,dp[i-1][-1j]有关。所以可以得到状态转移方程:dp[i][j] = 1 + MMIN(dp[i-1][j],dp[i][j-1],dp[i-1][j-1])
代码如下:
#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<queue>#include<vector>#include<algorithm>#include<map>#include<string>#include<set>#define LL long long#define MAX(a,b) (a>b?a:b)#define MIN(a,b) (a<b?a:b)#define MMIN(a,b,c) (MIN(a,b)<c?MIN(a,b):c)using namespace std;int a[1050][1050];int dp[1050][1050];int n,m;int ans = 0;int main(){ while(scanf("%d%d",&n,&m) && (m||n)) { ans = 0; for(int i = 1; i<=n; i++) for(int j = 1; j<=m; j++) { dp[i][j] = 0; scanf("%d",&a[i][j]); if(a[i][j]) { dp[i][j] = 1 + MMIN(dp[i-1][j],dp[i][j-1],dp[i-1][j-1]); } ans = MAX(dp[i][j],ans); } printf("%d\n",ans); } return 0;}
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