Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

In Ragannagar ( a small town in India), people are obsessed with palindromes . There are N road junctions(also callled points) labeled 0 to N-1 and roads exist between every pair of points. Roads are onewayed and for the road connecting point i to point j ( i < j) the direction to travel is i to j. Each road is labeled with a letter between 'A' to 'Z' . Rajar ,the traveler, wants to travel from point 0 to point N-1. However he wants to cover the longest palindromic path.

In the above arrangement the possible paths to take are:-

• ACCA
• ABA
• ACB
• BCA
• AD
• BB
• AA
• C

The largest palindrome amongst these is ACCA, so Rajar will take this path. Given the above configuration, help him decide which path to take.

Input

The first line of input will contain an integer denoting the number of test cases T <=25. Each test case will be formatted as follows:-

• The first line of each test case contains an integer denoting 2<= N <=50.
• The next N lines contain N characters each. Each character is a letter between 'A' to 'Z'. The jth character in the ith line denotes the label for the road between i to j and this will be equal to the ith character in the jth line.The ith character of the ith line will be * denoting no road exists.

Output

Output one line per case -
The longest palindromic path available or "NO PALINDROMIC PATH" if none exists.Note that quotes are for clarity only.
In case more than one longest path exists output the lexicographically smallest one.

Sample Input

2
5
*ABAC
A*CBD
BC*CB
ABC*A
CDBA*
5
*AXYZ
A*BQR
XB*BT
YQB*A
ZRTA*

Sample Output

ACCA
ABBA

---

題目要求從 0 走到 n-1 的最長回文路徑，並且輸出最小字典序的回文。

由於走法限定為 i->j (i < j) 時才會走，因此可以使用 dp 去解決。

dp[start][end] = max(dp[p][q]+2); if start->p ... q->end when p <= q

然後記錄最小值來自於哪，最後回溯處理即可。

因此是 O(n^4)。

#include <stdio.h>
#include <string.h>
#include <vector>
#include <queue>
using namespace std;
char g[105][105];
int dp[105][105];
int main() {
    int testcase, n;
    int i, j, k, p, q;
    scanf("%d", &testcase);
    while(testcase--) {
        scanf("%d", &n);
        for(i = 0; i < n; i++)
            scanf("%s", g[i]);
        memset(dp, 0, sizeof(dp));
        vector<int> dpArgv[105][105];
        for(i = 0; i < n; i++) {
            for(j = i+1; j < n; j++) {
                if(g[i][j] != '*') {
                    dp[i][j] = 1;
                    dpArgv[i][j].push_back((i<<10)+j);
                }
            }
        }
        for(i = 2; i < n; i++) {
            for(j = 0; i+j < n; j++) {
                //dp[j][i+j] j:start, i+j:end
                for(p = j+1; p < i+j; p++) { //j->p ... q->i+j
                    for(q = p; q < i+j; q++) {
                        if(g[j][p] == g[q][i+j] && g[j][p] != '*') {
                            if(dp[p][q]+2 >= dp[j][i+j]) {
                                if(dp[p][q]+2 > dp[j][i+j])
                                    dpArgv[j][i+j].clear();
                                dp[j][i+j] = dp[p][q]+2;
                                dpArgv[j][i+j].push_back((p<<10)+q);
                            }
                        }
                    }
                }
            }
        }
        if(dp[0][n-1] == 0) {
            puts("NO PALINDROMIC PATH");
            continue;
        }
        int p, q, tp, tq;
        queue<int> X, Y, rX, rY;
        X.push(0), Y.push(n-1);
        char ansBuf[105];
        int ansIdx = 0;
        while(!X.empty()) {
            char mn = 127;
            while(!X.empty()) {
                p = X.front(), X.pop();
                q = Y.front(), Y.pop();
                rX.push(p), rY.push(q);
                for(i = 0; i < dpArgv[p][q].size(); i++) {
                    tp = dpArgv[p][q][i]>>10;
                    tq = dpArgv[p][q][i]&1023;
                    if(p == tp) {
                        if(g[p][q] < mn)
                            mn = g[p][q];
                    } else if(g[p][tp] < mn)
                        mn = g[p][tp];
                }
            }
            ansBuf[ansIdx++] = mn;
            while(!rX.empty()) {
                p = rX.front(), rX.pop();
                q = rY.front(), rY.pop();
                for(i = 0; i < dpArgv[p][q].size(); i++) {
                    tp = dpArgv[p][q][i]>>10;
                    tq = dpArgv[p][q][i]&1023;
                    if(g[p][tp] == mn) {
                        if(tp != tq && p+1 != q)
                            X.push(tp), Y.push(tq);
                    }
                }
            }
        }
        if(dp[0][n-1]&1) {//odd
            for(i = 0; i < ansIdx; i++)
                putchar(ansBuf[i]);
            for(i = ansIdx-2; i >= 0; i--)
                putchar(ansBuf[i]);
        } else {
            for(i = 0; i < ansIdx; i++)
                putchar(ansBuf[i]);
            for(i = ansIdx-1; i >= 0; i--)
                putchar(ansBuf[i]);
        }
        puts("");
    }
    return 0;
}
/*
3
4
*AAA
A*AA
AA*A
AAA*
5
*ABAC
A*CBD
BC*CB
ABC*A
CDBA*
5
*AXYZ
A*BQR
XB*BT
YQB*A
ZRTA*
*/