Inna and Dima CodeForces - 374C

原题链接
考察:记忆化搜索
思路:
  不用等完整的DIMA在记录.设置\(dp[i][j]\)为当前坐标\((i,j)\)能走的步数.求所有起点的最大\(dp[i][j]\).
  无穷的条件是当前起点开始走到本次走过的地方.所以还需要一个\(bool\)数组记录.

Code

#include <iostream>
#include <cstring> 
using namespace std;
const int N = 1010,INF = 0x3f3f3f3f;
int n,m,dp[N][N];//以x,y为起点能搜到的最大dima 
char s[N][N],dir[N/10] = "DIMA";
bool vis[N][N];
int xx[4] = {-1,1,0,0},yy[4] = {0,0,-1,1};
int dfs(int x,int y,int idx)
{
	if(dp[x][y]!=-1) return dp[x][y];
	dp[x][y] = 0;
	int ans = 0;
	for(int i=0;i<4;i++)
	{
		int dx = x+xx[i],dy = y+yy[i];
		if(dx>0&&dx<=n&&dy>0&&dy<=m&&s[dx][dy]==dir[idx%4])
		{
			if(vis[dx][dy]) return INF; 
			vis[dx][dy] = 1;
			ans = max(dfs(dx,dy,idx+1),ans);
			vis[dx][dy] = 0;
		}
	}
	return dp[x][y] = ans+1;
}
void solve()
{
	int ans = 0;
	for(int i=1;i<=n;i++)
	  for(int j=1;j<=m;j++)
	   if(s[i][j]=='D'&&dp[i][j]==-1)
	   {
	   	vis[i][j] = 1;
	   	ans = max(dfs(i,j,1),ans);
	   	vis[i][j] = 0;
	   }
	if(ans/4==0) puts("Poor Dima!");
	else if(ans>=INF) puts("Poor Inna!");
	else printf("%d\n",ans/4);
}
int main()
{
	scanf("%d%d",&n,&m);
	memset(dp,-1,sizeof dp);
	for(int i=1;i<=n;i++) scanf("%s",s[i]+1);
	solve();
	return 0;
}