Largest Submatrix（動態規劃）

Largest Submatrix

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2018    Accepted Submission(s): 967

Problem Description

Now here is a matrix with letter 'a','b','c','w','x','y','z' and you can change 'w' to 'a' or 'b', change 'x' to 'b' or 'c', change 'y' to 'a' or 'c', and change 'z' to 'a', 'b' or 'c'. After you changed it, what's the largest submatrix with the same letters you can make?

 

 

Input

The input contains multiple test cases. Each test case begins with m and n (1 ≤ m, n ≤ 1000) on line. Then come the elements of a matrix in row-major order on m lines each with n letters. The input ends once EOF is met.

 

 

Output

For each test case, output one line containing the number of elements of the largest submatrix of all same letters.

 

 

Sample Input

2 4 abcw wxyz

 

 

Sample Output

3

題解：

上題 的增強版。

三種狀況。

所有變爲a,所有爲b，所有爲c,分別求最大。

代碼：

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<cmath>
 6 #include<vector>
 7 using namespace std;
 8 const int INF=0x3f3f3f3f;
 9 #define mem(x,y) memset(x,y,sizeof(x))
10 #define SI(x) scanf("%d",&x)
11 #define PI(x) printf("%d",x)
12 #define SD(x,y) scanf("%lf%lf",&x,&y)
13 #define P_ printf(" ")
14 const int MAXN=1010;
15 typedef long long LL;
16 int dp[3][MAXN][MAXN],s[MAXN],l[MAXN],r[MAXN];
17 char mp[MAXN][MAXN];
18 bool isa(char ch){
19     if(ch=='a'||ch=='w'||ch=='y'||ch=='z')
20         return true;
21     else return false;
22 }
23 bool isb(char ch){
24     if(ch=='b'||ch=='w'||ch=='x'||ch=='z')
25         return true;
26     else return false;
27 }
28 bool isc(char ch){
29     if(ch=='c'||ch=='x'||ch=='y'||ch=='z')
30         return true;
31     else return false;
32 }
33 
34 int main(){
35     int N,M;
36     while(~scanf("%d%d",&N,&M)){
37         for(int i=1;i<=N;i++)
38             scanf("%s",mp[i]+1);
39             mem(dp,0);
40             int ans=0;
41         for(int i=1;i<=N;i++){
42             for(int j=1;mp[i][j];j++){
43                 if(isa(mp[i][j]))dp[0][i][j]=dp[0][i-1][j]+1;
44                 s[j]=dp[0][i][j];l[j]=j;r[j]=j;
45             }
46             s[0]=s[M+1]=-1;
47             for(int j=1;j<=M;j++){
48                 while(s[l[j]-1]>=s[j])
49                     l[j]=l[l[j]-1];
50             }
51             for(int j=M;j>=1;j--){
52                 while(s[r[j]+1]>=s[j])
53                     r[j]=r[r[j]+1];
54             }
55             for(int j=1;j<=M;j++){
56                 ans=max((r[j]-l[j]+1)*s[j],ans);
57             }
58             //
59             for(int j=1;mp[i][j];j++){
60                 if(isb(mp[i][j]))dp[1][i][j]=dp[1][i-1][j]+1;
61                 s[j]=dp[1][i][j];l[j]=j;r[j]=j;
62             }
63             s[0]=s[M+1]=-1;
64             for(int j=1;j<=M;j++){
65                 while(s[l[j]-1]>=s[j])
66                     l[j]=l[l[j]-1];
67             }
68             for(int j=M;j>=1;j--){
69                 while(s[r[j]+1]>=s[j])
70                     r[j]=r[r[j]+1];
71             }
72             for(int j=1;j<=M;j++){
73                 ans=max((r[j]-l[j]+1)*s[j],ans);
74             }
75             //
76             for(int j=1;mp[i][j];j++){
77                 if(isc(mp[i][j]))dp[2][i][j]=dp[2][i-1][j]+1;
78                 s[j]=dp[2][i][j];l[j]=j;r[j]=j;
79             }
80             s[0]=s[M+1]=-1;
81             for(int j=1;j<=M;j++){
82                 while(s[l[j]-1]>=s[j])
83                     l[j]=l[l[j]-1];
84             }
85             for(int j=M;j>=1;j--){
86                 while(s[r[j]+1]>=s[j])
87                     r[j]=r[r[j]+1];
88             }
89             for(int j=1;j<=M;j++){
90                 ans=max((r[j]-l[j]+1)*s[j],ans);
91             }
92         }
93         printf("%d\n",ans);
94     }
95     return 0;
96 }