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

Problem C - Hippity Hopscotch

The game of hopscotch involves chalk, sidewalks, jumping, and picking things up. Our variant of hopscotch involves money as well. The game is played on a square grid of dimension n: each grid location is labelled (p,q) where 0 <= p < n and 0 <= q < n. Each grid location has on it a stack of between 0 and 100 pennies.

A contestant begins by standing at location (0,0). The contestant collects the money where he or she stands and then jumps either horizontally or vertically to another square. That square must be within the jumping capability of the contestant (say, k locations) and must have more pennies than those that were on the current square.

Given n, k, and the number of pennies at each grid location, compute the maximum number of pennies that the contestant can pick up before being unable to move.

Input Specification

The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.

• a line containing two integers between 1 and 100: n and k
• n lines, each with n numbers: the first line contains the number of pennies at locations (0,0) (0,1) ... (0,n-1); the next line contains the number of pennies at locations (1,0), (1,1), ... (1,n-1), and so on.

Output Specification

For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.

• a single integer giving the number of pennies collected

Sample Input

1

3 1
1 2 5
10 11 6
12 12 7

Output for Sample Input

37

---

由於有中文翻譯，就不多作說明。

題目解法：

很明顯地，由於只能往更高的數字爬，因此圖形是一個 DAG，那麼考慮將所有數字排序，
然後進行 DP 的更新，藉此得到最大值。

#include <stdio.h>
#include <algorithm>
using namespace std;
struct E {
    int i, j, v;
    bool operator<(const E &a) const {
        return v < a.v;
    }
};
int main() {
    E D[10005];
    int testcase;
    int i, j, k, x, y;
    int N, K;
    int g[105][105], dp[105][105];
    scanf("%d", &testcase);
    while(testcase--) {
        scanf("%d %d", &N, &K);
        int n = 0;
        for(i = 0; i < N; i++) {
            for(j = 0; j < N; j++) {
                scanf("%d", &g[i][j]);
                D[n].i = i, D[n].j = j;
                D[n].v = g[i][j];
                n++;
                dp[i][j] = 0;
            }
        }
        sort(D, D+n);
        dp[0][0] = g[0][0];
        int can[105][105] = {};
        for(i = 0; i < n; i++) {
            if(D[i].i == 0 && D[i].j == 0)
                break;
        }
        can[0][0] = 1;
        for(; i < n; i++) {
            if(can[D[i].i][D[i].j] == 0)    continue;
            j = 1, x = D[i].i+1, y = D[i].j;
            while(x < N && j <= K) {
                if(g[x][y] > D[i].v) {
                    dp[x][y] = max(dp[x][y], dp[D[i].i][D[i].j]+g[x][y]);
                    can[x][y] = 1;
                }
                x++, j++;
            }
            j = 1, x = D[i].i, y = D[i].j+1;
            while(y < N && j <= K) {
                if(g[x][y] > D[i].v) {
                    dp[x][y] = max(dp[x][y], dp[D[i].i][D[i].j]+g[x][y]);
                    can[x][y] = 1;
                }
                y++, j++;
            }
            j = 1, x = D[i].i-1, y = D[i].j;
            while(x >= 0 && j <= K) {
                if(g[x][y] > D[i].v) {
                    dp[x][y] = max(dp[x][y], dp[D[i].i][D[i].j]+g[x][y]);
                    can[x][y] = 1;
                }
                x--, j++;
            }
            j = 1, x = D[i].i, y = D[i].j-1;
            while(y >= 0 && j <= K) {
                if(g[x][y] > D[i].v) {
                    dp[x][y] = max(dp[x][y], dp[D[i].i][D[i].j]+g[x][y]);
                    can[x][y] = 1;
                }
                y--, j++;
            }
        }

        int ret = 0;
        for(i = 0; i < N; i++) {
            for(j = 0; j < N; j++) {
                ret = max(ret, dp[i][j]);
            }
        }
        printf("%d\n", ret);
        if(testcase)    puts("");
    }
    return 0;
}