hdu 1078（记忆化搜索）

传送门

题解：

简单记忆化搜索。

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;const int MAXN=104;int n,m;int a[MAXN][MAXN];int dp[MAXN][MAXN];int dx[4]={0,0,1,-1},dy[4]={1,-1,0,0};inline int read() {int x=0,f=1;char c=getchar();while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}while (c>='0'&&c<='9') x=x*10+c-'0',c=getchar();return x*f;}inline bool in(int a,int b) {return a>0&&a<=n&&b>0&&b<=n;}int dfs(int x,int y) {if (~dp[x][y]) return dp[x][y];int mx=0;for (int i=1;i<=m;++i)for (int j=0;j<4;++j) {int xx=x+dx[j]*i,yy=y+dy[j]*i;if (in(xx,yy)&&a[xx][yy]>a[x][y]) mx=max(mx,dfs(xx,yy));}return dp[x][y]=mx+a[x][y];}int main() {//freopen("hdu 1078.in","r",stdin);while (scanf("%d%d",&n,&m)&&~n&&~m) {memset(dp,-1,sizeof(dp));for (int i=1;i<=n;++i)for (int j=1;j<=n;++j)a[i][j]=read();printf("%d\n",dfs(1,1));}return 0;}