HDU 2859 Phalanx (DP)
Phalanx
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 363 Accepted Submission(s): 170
Problem Description
Today is army day, but the servicemen are busy with the phalanx for the celebration of the 60th anniversary of the PRC.
A phalanx is a matrix of size n*n, each element is a character (a~z or A~Z), standing for the military branch of the servicemen on that position.
For some special requirement it has to find out the size of the max symmetrical sub-array. And with no doubt, the Central Military Committee gave this task to ALPCs.
A symmetrical matrix is such a matrix that it is symmetrical by the “left-down to right-up” line. The element on the corresponding place should be the same. For example, here is a 3*3 symmetrical matrix:
cbx
cpb
zcc
A phalanx is a matrix of size n*n, each element is a character (a~z or A~Z), standing for the military branch of the servicemen on that position.
For some special requirement it has to find out the size of the max symmetrical sub-array. And with no doubt, the Central Military Committee gave this task to ALPCs.
A symmetrical matrix is such a matrix that it is symmetrical by the “left-down to right-up” line. The element on the corresponding place should be the same. For example, here is a 3*3 symmetrical matrix:
cbx
cpb
zcc
Input
There are several test cases in the input file. Each case starts with an integer n (0<n<=1000), followed by n lines which has n character. There won’t be any blank spaces between characters or the end of line. The input file is ended with a 0.
Output
Each test case output one line, the size of the maximum symmetrical sub- matrix.
Sample Input
3
abx
cyb
zca
4
zaba
cbab
abbc
cacq
0
Sample Output
3
3
Source
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gaojie
对于每个字符看该列以上和该行右侧的字符匹配量,如果匹配量大于右上角记录下来的矩阵大小,就是右上角的数值+1,否则就是这个匹配量。
#include <stdio.h> #include <algorithm> #include <iostream> #include <string.h> using namespace std; int dp[1010][1010]; char str[1010][1010]; int main() { int n; while(scanf("%d",&n) == 1 && n) { for(int i = 0;i < n;i++) scanf("%s",str[i]); int ans = 1; for(int i = 0;i < n;i++) for(int j = 0;j < n;j++) { if(i == 0 || j == n-1) { dp[i][j] = 1; continue; } int t1 = i, t2 = j; while(t1 >= 0 && t2 < n && str[t1][j] == str[i][t2]) { t1--; t2++; } t1 = i - t1; if(t1 >= dp[i-1][j+1]+1)dp[i][j] = dp[i-1][j+1]+1; else dp[i][j] = t1; ans = max(ans,dp[i][j]); } printf("%d\n",ans); } return 0; }
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