hdu1505
hdu1506的加强版， 如果要做这题，还是先去做1505吧
这一题，其实是对每一行做1505的那种dp，然后取最大值就行了。
1 #pragma warning(disable:4996) 2 #pragma comment(linker, "/STACK:1024000000,1024000000") 3 #include <stdio.h> 4 #include <string.h> 5 #include <time.h> 6 #include <math.h> 7 #include <map> 8 #include <set> 9 #include <queue> 10 #include <stack> 11 #include <vector> 12 #include <bitset> 13 #include <algorithm> 14 #include <iostream> 15 #include <string> 16 #include <functional> 17 #include <unordered_map> 18 typedef __int64 LL; 19 const int INF = 999999999; 20 21 22 const int N = 1000 + 10; 23 int a[N][N]; 24 int sum[N][N]; 25 int left[N], right[N]; 26 int main() 27 { 28 char str[11]; 29 int t, n, m; 30 scanf("%d", &t); 31 while (t--) 32 { 33 scanf("%d%d", &n, &m); 34 for (int i = 1;i <= n;++i) 35 { 36 for (int j = 1;j <= m;++j) 37 { 38 scanf("%s", str); 39 if (str[0] == 'R') 40 a[i][j] = 1; 41 else 42 a[i][j] = 0; 43 44 //得到每一行的高度 45 if (a[i][j] == 0) 46 sum[i][j] = sum[i - 1][j] + 1; 47 else 48 sum[i][j] = 0; 49 } 50 } 51 int ans = 0; 52 for (int i = 1;i <= n;++i) 53 { 54 left[1] = right[m] = 0; 55 int l, r; 56 for (int j = 2;j <= m;++j) 57 { 58 left[j] = 0; 59 l = j - 1; 60 while (l>=1 && sum[i][j] <= sum[i][l]) 61 { 62 left[j] += left[l] + 1; 63 l = l - left[l] - 1; 64 } 65 } 66 for (int j = m - 1;j >= 1;--j) 67 { 68 right[j] = 0; 69 r = j + 1; 70 while (r <= n && sum[i][j] <= sum[i][r]) 71 { 72 right[j] += right[r] + 1; 73 r = r + right[r] + 1; 74 } 75 } 76 for (int j = 1;j <= m;++j) 77 ans = std::max(ans, (left[j] + right[j] + 1)*sum[i][j]); 78 } 79 printf("%d\n", ans*3); 80 } 81 return 0; 82 }
