FatMouse and Cheese
Time Limit: 2000/1000 MS (java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 9050 Accepted Submission(s): 3792
PRoblem Description FatMouse has stored some cheese in a city. The city can be considered as a square grid of dimension n: each grid location is labelled (p,q) where 0 <= p < n and 0 <= q < n. At each grid location Fatmouse has hid between 0 and 100 blocks of cheese in a hole. Now he’s going to enjoy his favorite food.
FatMouse begins by standing at location (0,0). He eats up the cheese where he stands and then runs either horizontally or vertically to another location. The problem is that there is a super Cat named Top Killer sitting near his hole, so each time he can run at most k locations to get into the hole before being caught by Top Killer. What is worse – after eating up the cheese at one location, FatMouse gets fatter. So in order to gain enough energy for his next run, he has to run to a location which have more blocks of cheese than those that were at the current hole.
Given n, k, and the number of blocks of cheese at each grid location, compute the maximum amount of cheese FatMouse can eat before being unable to move.
Input There are several test cases. Each test case consists of
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 blocks of cheese at locations (0,0) (0,1) … (0,n-1); the next line contains the number of blocks of cheese at locations (1,0), (1,1), … (1,n-1), and so on. The input ends with a pair of -1’s.
Output For each test case output in a line the single integer giving the number of blocks of cheese collected.
Sample Input 3 1 1 2 5 10 11 6 12 12 7 -1 -1
Sample Output 37
代碼：
#include <iostream>#include <string>#include <cstring>#include <cstdio>#include <cmath>#include <cstdlib>#include <algorithm>#include <queue>#include <map>#define MST(s,q) memset(s,q,sizeof(s))#define INF 0x3f3f3f3f#define MAXN 1005using namespace std;int n, k, maxn;int Map[105][105];bool vis[105][105];int move_x[4] = {0, 0, 1, -1}, move_y[4] = {1, -1, 0, 0};int memory[105][105];int dfs(int x0, int y0){ if (memory[x0][y0]) return memory[x0][y0]; int ans = 0; for (int i = 0; i < 4; i++) for (int j = 1; j <= k; j++) { int x = x0 + j * move_x[i]; int y = y0 + j * move_y[i]; if (!vis[x][y] && Map[x][y] > Map[x0][y0] && x >= 0 && x < n && y >= 0 && y < n) { vis[x][y] = 1; int c = dfs(x, y); ans = c > ans ? c : ans; vis[x][y] = 0; } } return memory[x0][y0] = ans + Map[x0][y0];}int main(){ while (cin >> n >> k) { if (n == -1 && k == -1) break; MST(vis, 0);// 可省略 MST(memory, 0); for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) { scanf("%d", &Map[i][j]); } vis[0][0] = 1; printf("%d/n", dfs(0, 0) ); }}新聞熱點
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