演算法導論-15-6-在棋盤上移動

題目：

思考：

用一個矩陣s來儲存N*N的棋盤，用ans來儲存中間過程

（1）初始化：ans[i][j] = s[i][j] (i = N)

（2）遞推：ans[i][j] = MAX(s[i+1][j-1], s[i+1][j], s[i+1][j+1]) + s[i][j] (1<=i<N)

（3）結果：answer = MAX(s[i][j]) (i = 1)

代碼：

#include <iostream>using namespace std;#define N 10//為了便於處理邊界問題int map[N+2][N+2];int max(int a, int b, int c){if(a > b)return a > c ? a : c;elsereturn b > c ? b : c;}int main(){int i, j, n;while(cin>>n){memset(map, 0, sizeof(map));//產生隨機測試資料for(i = 1; i <= n; i++){for(j = 1; j <= n; j++){//cin>>map[i][j];map[i][j] = rand() % 10 - 5;cout<<map[i][j]<<' ';}cout<<endl;}int MAX = -0x7fffffff;//數組s與數組ans合并為數組map//ans[i][j] = s[i][j] (i = N)for(i = n-1; i >= 1; i--){for(j = 1; j <= n; j++){//ans[i][j] = MAX(s[i+1][j-1], s[i+1][j], s[i+1][j+1]) + s[i][j] (1<=i<N)map[i][j] = map[i][j] + max(map[i+1][j-1], map[i+1][j], map[i+1][j+1]);//answer = MAX(s[i][j]) (i = 1)if(i == 1 && map[i][j] > MAX)MAX = map[i][j];}}cout<<MAX<<endl;}return 0;}