FatMouse and Cheese
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.
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.
3 1 1 2 5 10 11 6 12 12 7 -1 -1
37
题意:给定一张N*N的图,每个格子上有一些豆子,初始位置是(0,0),每次只能向一个方向走最多m步,然后停下吃掉这个格子上的豆子,但有个限制是当前格子上的豆子必须比之前呆的一个格子里的豆子多,问最多能吃多少豆子。
题解:深搜,dp[i][j]表示从(i,j)点出发最多能吃到的豆子。
#include <stdio.h> #include <string.h> #define maxn 105 int G[maxn][maxn], n, m; int dp[maxn][maxn]; const int mov[][2] = {1, 0, -1, 0, 0, 1, 0, -1}; void getMap() { int i, j; for(i = 0; i < n; ++i) for(j = 0; j < n; ++j) scanf("%d", &G[i][j]); } int max(int a, int b) { return a > b ? a : b; } bool check(int x, int y) { if(x < 0 || x >= n || y < 0 || y >= n) return 0; return 1; } int DFS(int xx, int yy) { int i, j, x, y, maxv = 0; if(dp[xx][yy]) return dp[xx][yy]; for(i = 1; i <= m; ++i) { for(j = 0; j < 4; ++j) { x = xx + mov[j][0] * i; y = yy + mov[j][1] * i; if(check(x, y) && G[x][y] > G[xx][yy]) maxv = max(maxv, DFS(x, y)); } } return dp[xx][yy] = maxv + G[xx][yy]; } void solve() { int i, j; memset(dp, 0, sizeof(dp)); printf("%d\n", DFS(0, 0)); } int main() { // freopen("stdin.txt", "r", stdin); while(scanf("%d%d", &n, &m), n > 0) { getMap(); solve(); } return 0; }