Bargaining Table
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Bob wants to put a new bargaining table in his office. To do so he measured the office room thoroughly and drew its plan: Bob's office room is a rectangular room n × m meters. Each square meter of the room is either occupied by some furniture, or free. A bargaining table is rectangular, and should be placed so, that its sides are parallel to the office walls. Bob doesn't want to change or rearrange anything, that's why all the squares that will be occupied by the table should be initially free. Bob wants the new table to sit as many people as possible, thus its perimeter should be maximal. Help Bob find out the maximum possible perimeter of a bargaining table for his office.
The first line contains 2 space-separated numbers n and m (1 ≤ n, m ≤ 25) — the office room dimensions. Then there follow n lines with m characters 0 or 1 each. 0 stands for a free square meter of the office room. 1 stands for an occupied square meter. It's guaranteed that at least one square meter in the room is free.
Output one number — the maximum possible perimeter of a bargaining table for Bob's office room.
3 3000010000
8
5 411000000000000000000
16
思路:
这道题可能有很多方法,下面为其中之一。
将每种情况的值存到MAP四维数组里,同时比较每种情况可以坐的人数,最后将最大值输出即可。
代码:
#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int Map[30][30];int ans(int i,int k,int j,int l)//每种情况下的值,如果为0则表明全为空地,若大于0则说明此情况有非空地存在。{ int p,sum=0,q; for(p=i;p<=k;p++) for(q=j;q<=l;q++) sum+=Map[p][q]; return sum;}int main(){ int n,m,MAP[26][26][26][26],i,k,j,l; char a[100]; while(~scanf("%d%d",&n,&m)) { for(i=0;i<n;i++) { scanf("%s",a); for(j=0;j<m;j++) Map[i][j]=a[j]-'0';//空地为0,非空地为1。 } int answer=0; for(i=0;i<n;i++) for(k=i;k<n;k++) for(j=0;j<m;j++) for(l=j;l<m;l++) { MAP[i][k][j][l]=ans(i,k,j,l); if(MAP[i][k][j][l]==0)//全是空地的情况; { answer=max(answer,(k-i+1+l-j+1)*2);//可以坐的人数的最大值; } } printf("%d\n",answer); }}
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