UVA 10285 Longest Run on a Snowboard

标签：滑雪   dp   动态规划   dfs   记忆化搜索   

记忆化搜索，跟以前的做过的 滑雪 一样的。

DP+DFS。用dp[][]保存搜索记录，然后满足条件累加即可。

#include<cstdio>
#include<cstring>
#include<string>
#include<queue>
#include<algorithm>
#include<map>
#include<stack>
#include<iostream>
#include<list>
#include<set>
#include<bitset>
#include<vector>
#include<cmath>

#define INF 0x7fffffff
#define eps 1e-8
#define LL long long
#define PI 3.141592654
#define CLR(a,b) memset(a,b,sizeof(a))
#define FOR(i,a,b) for(int i=a;i<b;i++)
#define FOR_(i,a,b) for(int i=a;i>=b;i--)
#define pub push_back
#define puf push_front
#define pob pop_back
#define pof pop_front
#define mp make_pair
#define ft first
#define sd second
#define sf scanf
#define pf printf
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(),(v).end()
#define acfun std::ios::sync_with_stdio(false)

#define SIZE 100 +1
using namespace std;

int n,m;
int g[SIZE][SIZE];
int xx[]={0,0,-1,1};
int yy[]={-1,1,0,0};
int dp[SIZE][SIZE];

int dfs(int i,int j)
{
    if(dp[i][j]>1)return dp[i][j];
    FOR(k,0,4)
    {
        int x=i+xx[k];
        int y=j+yy[k];
        if(x<0||y<0||x>=n||y>=m)continue;
        if(g[x][y]<g[i][j])
            dp[i][j]=max(dp[i][j],dfs(x,y)+1);
    }
    return dp[i][j];
}

int main()
{
    int t;
    sf("%d",&t);
    while(t--)
    {
        char str[SIZE];
        sf("%s%d%d",str,&n,&m);
        FOR(i,0,n)
        FOR(j,0,m)
        sf("%d",&g[i][j]),dp[i][j]=1;

        int ans=0;
        FOR(i,0,n)
        FOR(j,0,m)
         ans=max(ans,dfs(i,j));
        pf("%s: %d\n",str,ans);
    }
}

UVA 10285 Longest Run on a Snowboard

标签：滑雪   dp   动态规划   dfs   记忆化搜索   

原文地址：http://blog.csdn.net/dongshimou/article/details/41442813