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FatMouse and Cheese
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10863 Accepted Submission(s): 4625
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
#include <iostream>
#include <cstdio>
#include <cstring>
#include <ctime>
#include <algorithm>
#include <cmath>
using namespace std;
const int maxn =105;
int a[maxn][maxn],dp[maxn][maxn];
int n,k;
int dx[]={0,0,1,-1};
int dy[]={1,-1,0,0};
int dfs(int x, int y){
int tmp=0;
for (int j=1;j<=k;j++){
for (int i=0;i<4;i++){
int xx=x+dx[i]*j; //k步是水平或竖直的
int yy=y+dy[i]*j;
if (xx<0 || yy<0 || xx>=n || yy>=n) continue;
if (a[x][y]>=a[xx][yy]) continue;
if (dp[xx][yy]) { //如果已经探索过了，就没必要继续深搜了
tmp=max(tmp,dp[xx][yy]); //取最大值
continue;
}
tmp=max(tmp,dfs(xx,yy)); //取所有方案中的最大值
}
}
return dp[x][y]=a[x][y]+tmp;
}
int main(){
std::ios::sync_with_stdio(false); //提高cin输入效率
while (cin >> n >> k){
if (n==-1 && k == -1) break;
for (int i=0;i<n;i++){
for (int j=0;j<n;j++) cin>>a[i][j];
}
memset(dp,0,sizeof(dp));
cout<<dfs(0,0)<<endl; //最优解在出发点上
}
return 0;
}
发布者：全栈程序员-站长，转载请注明出处：https://javaforall.net/164447.html原文链接：https://javaforall.net
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