HDU-2870 Largest Submatrix (线性dp 最大01矩阵)(2009 Multi-University Training Contest 7 )
2017-11-16 15:52
483 查看
Largest Submatrix
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2656 Accepted Submission(s): 1298
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
#include <bits/stdc++.h>
using namespace std;
int R, C, a[1001][1001], h[1001][1001], l[1001], r[1001];
char s[1001][1001];
int solve(int kind){
for(int i = 1; i <= R; ++i){
for(int j = 1; j <= C; ++j){
if(kind == 0){
if(s[i][j] == 'a' || s[i][j] == 'w' || s[i][j] == 'y' || s[i][j] == 'z') a[i][j] = 1;
else a[i][j] = 0;
}
if(kind == 1){
if(s[i][j] == 'b' || s[i][j] == 'w' || s[i][j] == 'x' || s[i][j] == 'z') a[i][j] = 1;
else a[i][j] = 0;
}
if(kind == 2){
if(s[i][j] == 'c' || s[i][j] == 'x' || s[i][j] == 'y' || s[i][j] == 'z') a[i][j] = 1;
else a[i][j] = 0;
}
}
}
for(int i = 1; i <= C; ++i){
h[0][i] = 0;
}
int ans = 0;
for(int i = 1; i <= R; ++i){
for(int j = 1; j <= C; ++j){
if(a[i][j] == 1){
h[i][j] = h[i - 1][j] + 1;
}
else{
h[i][j] = 0;
}
}
}
for(int i = 1; i <= R; ++i){
for(int j = 1; j <= C; ++j){
l[j] = r[j] = j;
}
h[i][0] = -1;
for(int j = 1; j <= C; ++j){
while(h[i][j] <= h[i][l[j] - 1]){
l[j] = l[l[j] - 1];
}
}
h[i][C + 1] = -1;
for(int j = C - 1; j >= 1; --j){
while(h[i][j] <= h[i][r[j] + 1]){
r[j] = r[r[j] + 1];
}
}
for(int j = 1; j <= C; ++j){
ans = max(ans, (r[j] - l[j] + 1) * h[i][j]);
}
}
return ans;
}
int main(){
while(scanf("%d %d", &R, &C) != EOF){
for(int i = 1; i <= R; ++i){
for(int j = 1; j <= C; ++j){
s[i][j] = getchar();
while(s[i][j] < 'a' || s[i][j] > 'z'){
s[i][j] = getchar();
}
}
}
int ans = 0;
ans = max(ans, solve(0));
ans = max(ans, solve(1));
ans = max(ans, solve(2));
printf("%d\n", ans);
}
}
/*
题意:
1000*1000的字符矩阵,字符只包含a,b,c,w,x,y,z,w,x,y,z可以做一些变化变成a或b或c,问最大只包含一种字符的
矩阵面积。
思路:
首先有一个观察,我们不用考虑包含w,x,y,z的矩阵情况,将他们转化成a或b或c处理即可。如果直接考虑怎么处理这些变化
然后计算矩阵面积比较麻烦,我们不妨分开处理,分别计算只包含a的情况,然后b,c,最后取最大值即可。
计算只包含a的情况那么就可以转换成01矩阵中最大1矩阵的问题。可以参考: http://blog.csdn.net/hmc0411/article/details/78540500; http://blog.csdn.net/hmc0411/article/details/78539159;
*/
相关文章推荐
- (HDU 5823)2016 Multi-University Training Contest 8 color II (m染色问题、最大独立集、DP)
- 2013 Multi-University Training Contest 9(hdu 4686 - 4691)dp(好)+矩阵快速幂+一般图匹配带花树+后缀数组
- HDU 2870 Largest Submatrix(dp最大子矩阵和)
- HDU 2870 Largest Submatrix DP求最大子矩阵
- HDU2870_Largest Submatrix【最大完全子矩阵】
- HDU 2870 Largest Submatrix(最大子矩阵面积)
- HDU 2870 Largest Submatrix(最大完全子矩阵之不可移动列)
- Largest Submatrix-最大子矩阵-HDU-2870
- 2015 Multi-University Training Contest 10 (hdu 5406-5416)数据结构+dp+矩阵快速幂+bitset优化拓扑排序+(dp&树状数组)
- hdu 2818 Building Block(加权并查集)2009 Multi-University Training Contest 1
- HDU/HDOJ 2855 2009 Multi-University Training Contest 5 - Host by NUDT 矩阵二分幂
- hdu 3094 A tree game 2009 Multi-University Training Contest 18 - Host by ECNU
- HDU-2870-Largest Submatrix(DP)
- hdu 4370 0 or 1 (最短路 // 01 规划 2012 Multi-University Training Contest 8 )
- hdu 5726 2016 Multi-University Training Contest 1(二分+dp)
- hdu 2870 Largest Submatrix(dp)
- 2017 Multi-University Training Contest 10 1002 Array Challenge HDU 6172(找规律 矩阵快速幂)
- HDU 4322 最大费用最大流 2012 Multi-University Training Contest 3
- 2017 Multi-University Training Contest - Team 1 1003&&HDU 6035 Colorful Tree【树形dp】
- HDU 2870 Largest Submatrix (最大子矩形面积)