Example 1:
Input:
1 2
01
Output:
0
Example Two:
Input:
1 3
101
Output:
1
Example Three (with their own self-test to find the error):
Input:
6 6
100111
111011
111111
111111
111111
101111
Output:
16
Meaning of the questions: find a second large rectangular made only by 1 composed of rectangular consisting of 1 and 0. (Prefix and thinking)
Problem-solving ideas: (1) this solution is to look at someone else's code after the game, this game also has its own ideas, but did not thoughtful. Thought the use of prefixes and to the calculated position in front of each of the successive locations of a total number of 1, and then began to count down from the top left corner, and the length of each of the current position (rear position used to calculate the height of the rectangle) records, each calculate the size of the current times the size of the largest and the second largest rectangle rectangular, update the answer.
(2) dfs, pruning feel not good enough so they time out, not figured out how not to pruning methods timeout, first keep, complexity is O (N³), by 95%.
Do not say, paste the code:
(1) AC Code
1 #include<bits/stdc++.h> 2 using namespace std; 3 const int maxn=1e3+20; 4 int a[maxn][maxn],max_1=0,max_2=0;///a数组记录每个位置的前缀和，max_1和max_2为最大和次大 5 char ss[maxn][maxn]; 6 struct node 7 { 8 int num,ins; 9 }pp[maxn];///记录当前位置的长度和位置 10 void update(int summ)///更新最大值和次大值 11 { 12 if(summ>max_1){ 13 swap(max_1,max_2);swap(summ,max_1); 14 } 15 else if(summ>max_2)swap(summ,max_2); 16 } 17 int main() 18 { 19 int n,m; 20 scanf("%d%d",&n,&m); 21 for(int i=1; i<=n; i++) 22 scanf("%s",ss[i]); 23 for(int i=1;i<=n;i++){ 24 for(int j=0;j<m;j++){ 25 if(ss[i][j]=='0')a[i][j+1]==0; 26 else if(ss[i][j]=='1')a[i][j+1]=a[i][j]+1; 27 } 28 } 29 for(int j=1; j<=m; j++) 30 { 31 int r=0; 32 for(int i=1; i<=n; i++) 33 { 34 int book=i; 35 while(r&&pp[r].num>a[i][j]){ 36 update((i-pp[r].ins)*pp[r].num); 37 update((i-pp[r].ins-1)*pp[r].num); 38 book=min(book,pp[r].ins); 39 r--; 40 } 41 pp[++r]={a[i][j],book}; 42 } 43 while(r){ 44 update(((n+1)-pp[r].ins)*pp[r].num); 45 update(((n+1)-pp[r].ins-1)*pp[r].num); 46 r--; 47 } 48 } 49 printf("%d\n",max_2); 50 return 0; 51 }
（2）有剪枝通过的大佬教我一下
1 #include<bits/stdc++.h> 2 using namespace std; 3 const int maxn=1e3+20; 4 int a[maxn][maxn],max_1=0,max_2=0,t;///a数组记录每个位置的前缀和，max_1和max_2为最大和次大 5 int jlen=0,ilen=0,rec;///用于记录最小长度，最高高度和当前面积 6 char ss[maxn][maxn]; 7 void dfs(int x,int y) 8 { 9 if(a[x][y]) 10 { 11 if(a[x][y]>=jlen) 12 { 13 ilen++; 14 rec=ilen*jlen; 15 if(rec>max_1){ 16 swap(max_1,max_2);swap(rec,max_1);} 17 else if(rec>max_2) 18 swap(rec,max_2); 19 dfs(x-1,y); 20 } 21 else if(a[x][y]<jlen) 22 { 23 ilen++,jlen=a[x][y]; 24 rec=ilen*jlen; 25 if(rec>max_1) 26 t=max_1,max_1=rec,max_2=t; 27 else if(rec>max_2) 28 max_2=rec; 29 dfs(x-1,y); 30 } 31 } 32 return; 33 } 34 int main() 35 { 36 int n,m,x,book=0; 37 scanf("%d%d",&n,&m); 38 for(int i=1; i<=n; i++){ 39 scanf("%s",ss[i]); 40 for(int j=0; j<m; j++) 41 { 42 if(j==0&&ss[i][0]=='1') 43 a[i][1]=1,book++; 44 else if(ss[i][j]=='0') 45 a[i][j+1]=0; 46 else if(ss[i][j]=='1') 47 a[i][j+1]=a[i][j]+1,book++; 48 } 49 } 50 for(int i=1; i<=n; i++) 51 { 52 for(int j=1; j<=m; j++) 53 { 54 if(a[i][j]) 55 { 56 ilen=1,jlen=a[i][j]; 57 rec=ilen*jlen; 58 if(rec>max_1) 59 t=max_1,max_1=rec,max_2=t; 60 else if(rec>max_2) 61 max_2=rec; 62 dfs(i-1,j); 63 } 64 } 65 } 66 printf("%d\n",max_2); 67 return 0; 68 }