HDU 2870 Largest Submatrix (最大子矩阵)

来源：互联网发布：如何做数据统计编辑：程序博客网时间：2024/05/22 00:20

Largest Submatrix

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 860 Accepted Submission(s): 421

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

View Code

  1 /*  2   跟1505类似，只不过要加上枚举，比较麻烦一点而已。  3   首先将可变成a的全部变成a，记录每一层中a左右可达的边界，然后从上到下找出第i层往上能达到的最大矩阵  4   然后同样把可变成b的全部变成b，做同样的步骤，最后再枚举c。求出最大值即使答案。  5   6   375MS    13160K  7   8 #include <iostream>  9 #include <cstdio> 10 #include <cstring> 11 #include <string> 12 #include <algorithm> 13 #define SIZE 1005 14  15 using namespace std; 16  17 int lt[SIZE][SIZE],rt[SIZE][SIZE]; 18 int height[SIZE][SIZE]; 19 int n,m; 20 int ans; 21 string s[SIZE]; 22  23 void cal() 24 { 25     for(int i=1; i<=n; i++) 26         for(int j=1; j<=m; j++) 27             lt[i][j] = rt[i][j] = j; 28     for(int i=1; i<=n; i++) 29         height[i][0] = height[i][m+1] = -1; 30     for(int i=1; i<=n; i++) 31     { 32         for(int j=1; j<=m; j++) 33         { 34             while(height[i][lt[i][j]-1] >= height[i][j]) 35                 lt[i][j] = lt[i][lt[i][j]-1]; 36         } 37     } 38     for(int i=1; i<=n; i++) 39     { 40         for(int j=m; j>=1; j--) 41         { 42             while(height[i][rt[i][j]+1] >= height[i][j]) 43                 rt[i][j] = rt[i][rt[i][j]+1]; 44         } 45     } 46     for(int i=1; i<=n; i++) 47         for(int j=1; j<=m; j++) 48             ans = max(ans,(rt[i][j]-lt[i][j]+1)*height[i][j]); 49 } 50  51 int main() 52 { 53     while(~scanf("%d%d",&n,&m)) 54     { 55         for(int i=1; i<=n; i++) 56             s[i].clear(); 57         string temp; 58         for(int i=1; i<=n; i++) 59         { 60             cin >> temp; 61             s[i] += '0'; 62             s[i] += temp; 63         } 64         ans = 0; 65         for(int i=0; i<=m; i++) 66             height[0][i] = 0; 67         for(int i=1; i<=n; i++) 68         { 69             for(int j=1; j<=m; j++) 70             { 71                 if(s[i][j] == 'a' || s[i][j] == 'w' || s[i][j] == 'y' || s[i][j] == 'z') 72                     height[i][j] = height[i-1][j]+1; 73                 else 74                     height[i][j] = 0; 75             } 76         } 77         cal(); 78         for(int i=1; i<=n; i++) 79         { 80             for(int j=1; j<=m; j++) 81             { 82                 if(s[i][j] == 'b' || s[i][j] == 'w' || s[i][j] == 'x' || s[i][j] == 'z') 83                      height[i][j] = height[i-1][j]+1; 84                 else 85                     height[i][j] = 0; 86             } 87         } 88         cal(); 89         for(int i=1; i<=n; i++) 90         { 91             for(int j=1; j<=m; j++) 92             { 93                 if(s[i][j] == 'c' || s[i][j] == 'y' || s[i][j] == 'x' || s[i][j] == 'z') 94                      height[i][j] = height[i-1][j]+1; 95                 else 96                     height[i][j] = 0; 97             } 98         } 99         cal();100         printf("%d\n",ans);101     }102     return 0;103 }104 */105 106 /*107   可将lt[][],rt[][]降维成lt[],rt[]，循环每一层的时候顺便求出最大值，节省了一半空间消耗108   343MS    5272K109   */110 #include <iostream>111 #include <cstdio>112 #include <cstring>113 #include <string>114 #include <algorithm>115 #define SIZE 1005116 117 using namespace std;118 119 int lt[SIZE],rt[SIZE];120 int height[SIZE][SIZE];121 int n,m;122 int ans;123 string s[SIZE];124 125 void cal()126 {127     for(int i=1; i<=n; i++)128     {129         for(int j=1; j<=m; j++)130             lt[j] = rt[j] = j;131         height[i][0] = height[i][m+1] = -1;132         for(int j=1; j<=m; j++)133         {134             while(height[i][lt[j]-1] >= height[i][j])135                 lt[j] = lt[lt[j]-1];136         }137         for(int j=m; j>=1; j--)138         {139             while(height[i][rt[j]+1] >= height[i][j])140                 rt[j] = rt[rt[j]+1];141         }142         for(int j=1; j<=m; j++)143             ans = max(ans,(rt[j]-lt[j]+1)*height[i][j]);144     }145 }146 147 int main()148 {149     while(~scanf("%d%d",&n,&m))150     {151         for(int i=1; i<=n; i++)152             s[i].clear();153         string temp;154         for(int i=1; i<=n; i++)155         {156             cin >> temp;157             s[i] += '0';158             s[i] += temp;159         }160         ans = 0;161         for(int i=0; i<=m; i++)162             height[0][i] = 0;163         for(int i=1; i<=n; i++)164         {165             for(int j=1; j<=m; j++)166             {167                 if(s[i][j] == 'a' || s[i][j] == 'w' || s[i][j] == 'y' || s[i][j] == 'z')168                     height[i][j] = height[i-1][j]+1;169                 else170                     height[i][j] = 0;171             }172         }173         cal();174         for(int i=1; i<=n; i++)175         {176             for(int j=1; j<=m; j++)177             {178                 if(s[i][j] == 'b' || s[i][j] == 'w' || s[i][j] == 'x' || s[i][j] == 'z')179                      height[i][j] = height[i-1][j]+1;180                 else181                     height[i][j] = 0;182             }183         }184         cal();185         for(int i=1; i<=n; i++)186         {187             for(int j=1; j<=m; j++)188             {189                 if(s[i][j] == 'c' || s[i][j] == 'y' || s[i][j] == 'x' || s[i][j] == 'z')190                      height[i][j] = height[i-1][j]+1;191                 else192                     height[i][j] = 0;193             }194         }195         cal();196         printf("%d\n",ans);197     }198     return 0;199 }