POJ 1088 滑雪 记忆化搜索

解析：状态d[i][j]代表r=i , c=j这个位置能滑的最大长度。深搜+备忘录

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

const int maxn=100+5;
int R,C;
int a[maxn][maxn];
int d[maxn][maxn];
int dr[]={0,-1,0,1};
int dc[]={-1,0,1,0};

int solve(int r,int c)
{
    if(d[r][c]>0) return d[r][c];
    int next_r,next_c;
    int maxv=0;
    for(int i=0;i<4;i++)
    {
        next_r=r+dr[i];
        next_c=c+dc[i];
        if(next_r<=0 || next_r>R || next_c<=0 || next_c>C)
            continue;
        if(a[r][c]>a[next_r][next_c])
            maxv=max(maxv,solve(next_r,next_c));
    }
    return d[r][c]=maxv+1;
}
int main()
{
    //freopen("in.txt","r",stdin);
    while(cin>>R>>C)
    {
        for(int i=1; i<=R; i++)
            for(int j=1; j<=C; j++)
                scanf("%d",&a[i][j]);
        memset(d,0,sizeof(d));
        int ans=0;
        for(int i=1;i<=R;i++)
            for(int j=1;j<=C;j++)
                ans=max(ans,solve(i,j));
        printf("%d
",ans);
    }
    return 0;
}