FatMouse and Cheese(记忆化搜索)
来源:互联网 发布:蓝月翅膀升级数据 编辑:程序博客网 时间:2024/06/07 13:36
FatMouse and Cheese
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 10857 Accepted Submission(s): 4622
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.
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.
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 11 2 510 11 612 12 7-1 -1
Sample Output
37
Source
Zhejiang University Training Contest 2001
Recommend
记忆化搜索不过最多可以走k步
记忆化搜索不过最多可以走k步
#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <queue>#define inf 110using namespace std;int a[inf][inf], dp[inf][inf];int yi[4][2] = { {1, 0}, {-1, 0}, {0, 1}, {0, -1}};int DFS(int x, int y);int n, k;int judge(int x, int y){ if(x < 1 || x > n || y < 1 || y > n) return 1; return 0;}int main(){ while(~scanf("%d %d", &n, &k)) { if(n < 1 || k < 1) break; memset(dp, 0, sizeof(dp)); memset(a, 0, sizeof(a)); for(int i = 1; i <= n; i++) for(int j = 1; j <= n; j++) scanf("%d", &a[i][j]); int maxs = DFS(1, 1); printf("%d\n", maxs); } return 0;}int DFS(int x, int y){ int xx, yy; if(!dp[x][y]) { int ans = 0; for(int i = 1; i <= k; i++) for(int j = 0; j < 4; j++) { xx = x + yi[j][0] * i; yy = y + yi[j][1] * i; if(judge(xx, yy)) continue; else if(a[xx][yy] > a[x][y]) ans = max(ans, DFS(xx, yy)); } dp[x][y] = ans + a[x][y]; } return dp[x][y];}
阅读全文
0 0
- hdu1078 FatMouse and Cheese(记忆化搜索)
- HDU1078 FatMouse and Cheese 【记忆化搜索】
- hdoj1078 FatMouse and Cheese【记忆化搜索】
- hdu1078 FatMouse and Cheese 记忆化搜索
- HDU1078 FatMouse and Cheese记忆化搜索
- hdoj1078 FatMouse and Cheese(记忆化搜索)
- FatMouse and Cheese(记忆化搜索)
- hdu1078 fatmouse and cheese 记忆化搜索
- hdu1078 FatMouse and Cheese【记忆化搜索】
- hdu 1078 fatmouse and cheese 记忆化搜索
- hdu 1078 FatMouse and Cheese(深搜----记忆化搜索)
- hdu 1078 FatMouse and Cheese(dp 记忆化搜索)
- hdu 1078 FatMouse and Cheese(记忆化搜索)
- ZOJ 1107 FatMouse and Cheese(记忆化搜索)
- hdoj 1078 FatMouse and Cheese(记忆化搜索)
- hdu 1078 FatMouse and Cheese 记忆化搜索
- hdu 1078 FatMouse and Cheese (记忆化搜索 )
- HDU 1078 FatMouse and Cheese(记忆化搜索)
- 如何解决WIN8.1USB转串口不能识别设备,代码10
- 问题 : 数圈
- QNX学习笔记——IPC(5)
- P
- 如何设置jvm内存
- FatMouse and Cheese(记忆化搜索)
- linux 如何清理僵尸进程
- minimum-depth-of-binary-tree
- 通知栏通知,及点击跳转(广播实现)
- 《算法导论》第四章-思考题(参考答案)
- QNX手册学习笔记————中断机制
- TensorFlow手写汉字识别
- python numpy
- http协议几种数据传输方式