FatMouse and Cheese
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Description
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
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
Sample Input
1 2 5
10 11 6
12 12 7
-1 -1
Sample Output
/*
题意:给你一个n*n的矩阵,每个点都有一定数量的食物,出发点在(0,0),每次可以走最多k步,下一个单元格的食物数量
一定要比当前单元格的数量多,问你最多能吃到的食物数量 初步思路:记忆化搜索
*/
#include <bits/stdc++.h>
#define N 110
using namespace std;
int n,k;
int mapn[N][N];
int dp[N][N];//dp[i][j]用于保存到达i j之后能吃到的最大食物数量
int dir[][]={{,},{-,},{,},{,-}};
bool vis[N][N];
bool ok(int x,int y,int val){
if(x<||x>=n||y<||y>=n||vis[x][y]||mapn[x][y]<=val) return false;
return true;
}
int dfs(int x,int y){//三种 状态坐标,第几步
if(dp[x][y]!=-) //遇到合适的状态直接返回就行了
return dp[x][y];
int cur=;
for(int i=;i<;i++){
int fx=x,fy=y;
for(int j=;j<=k;j++){
fx+=dir[i][];
fy+=dir[i][];
if(ok(fx,fy,mapn[x][y])==false) continue; vis[fx][fy]=true; cur=max(cur,dfs(fx,fy)); vis[fx][fy]=false; //一条路走完了,将标记清理干净
}
}
return dp[x][y]=cur+mapn[x][y];
}
void init(){
memset(dp,-,sizeof dp);
memset(vis,false,sizeof vis);
}
int main(){
// freopen("in.txt","r",stdin);
while(scanf("%d%d",&n,&k)!=EOF&&(n!=-&&k!=-)){
init();
for(int i=;i<n;i++){
for(int j=;j<n;j++){
scanf("%d",&mapn[i][j]);
}
}//处理输入
dfs(,);
// for(int i=0;i<n;i++){
// for(int j=0;j<n;j++){
// cout<<dp[i][j]<<" ";
// }
// cout<<endl;
// }
printf("%d\n",dp[][]);
}
return ;
}
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