标签：list break mint struct ecif empty names using problem
SOJ 3329: Maximum Submatrix II http://acm.scu.edu.cn/soj/problem.action?id=3329
Problem: Given a $0-1$ matrix, find the maximum submatrix which contains only 0s and output the number of 0s in it.
Technique:
(1) Dynamic Programming
Denote $dp[i][j], 1\le i\le n, 1\le j\le m$ as the maximum number of sequential 0s ended at $[i,j]$, then the number of 0s in the maximum submatrix ended at $[i,j]$ is
$\max_{i‘‘\le i‘\le i}\min_{i‘\le k\le i}dp[k][j]*(i-k+1)$.
The time complexity is $O(m^{2}n)$.
Code:
#include<iostream>
using namespace std;
int mat[105][105];
int dp[105][105];
int main()
{
int n,m;
int T;
int i,j,k;
int temp;
int minTep;
int ans;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
for(j=0;j<=m;j++)
{
mat[0][j]=1;
dp[0][j]=0;
}
for(i=0;i<=n;i++)
{
mat[i][0]=1;
dp[i][0]=0;
}
for(i=1;i<=n;i++)
for(j=1;j<=m;j++)
{
scanf("%d",&mat[i][j]);
dp[i][j]=mat[i][j] ? 0 : dp[i][j-1]+1;
}
ans=0;
for(i=1;i<=n;i++)
for(j=1;j<=m;j++)
{
temp=0;
minTep=m+1;
for(k=i;k>0;k--)
{
if(dp[k][j])
{
minTep=dp[k][j]<minTep ? dp[k][j] : minTep;
temp=temp>minTep*(i-k+1) ? temp : minTep*(i-k+1);
}
else
break;
}
ans=temp>ans ? temp : ans;
}
printf("%d\n",ans);
}
return 0;
}
(2) Monotonic Stack
Note that given a list of $dp[k][j], i‘\le k\le i$, the number of 0s in the submatrix is the minimum of $dp[k][j]$ times the length of the list, which is similar to SOJ 3085 (see here). Specifically, we can maintain an increasing stack.
0 0 0 $dp[i‘][j]$
0 0 $\dots$
0 $dp[i-1][j]$
0 0 0 0 $dp[i][j]$
Code:
#include<iostream> #include<stack> using namespace std; struct node { int num; int no; int prevNo; }; int mat[1005][1005]; int main() { int n,m; int i,j; stack<node>s; int temp; int ans; node x; int T; scanf("%d",&T); while(T--) { scanf("%d%d",&n,&m); for(j=1;j<=m;j++) { mat[0][j]=0; mat[n+1][j]=0; } for(i=1;i<=n;i++) mat[i][0]=0; for(i=1;i<=n;i++) for(j=1;j<=m;j++) { scanf("%d",&temp); mat[i][j]=temp ? 0 : mat[i][j-1]+1; } ans=0; for(j=1;j<=m;j++) for(i=0;i<=n+1;i++) { while(!s.empty() && mat[i][j]<s.top().num) { temp=s.top().num*(i-1-s.top().prevNo); ans=temp>ans ? temp : ans; s.pop(); } x.num=mat[i][j]; x.no=i; x.prevNo=s.size() ? s.top().no : 0; s.push(x); } while(!s.empty()) s.pop(); printf("%d\n",ans); } return 0; }
标签：list break mint struct ecif empty names using problem
原文地址：https://www.cnblogs.com/ClearMoonlight/p/10540508.html
