1026-windy数+数位DP+记忆化搜索

1026: [SCOI2009]windy数

题意：数位DP模板题；

目前只理解了记忆化搜索，就想练练手，

------给递推写法留一个位子

------

　　注意这道题要判断前导0的情况，1 ）可以加一个bool lead，或者在（i==0&&pre==-10)特判

#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <cstdio>
using namespace std;
typedef long long ll;
ll dp[10][10],digit[10];

ll dfs(int pos,int pre,bool limit,bool lead)//lead是判断前导0的，如果（i==0&&pre==-10）需要在dfs中继续pre=-10，避免满足记忆化的条件
{
    if(pos==0)return 1;
    if(!limit&&dp[pos][pre]>=0&&!lead)return dp[pos][pre];
    int num = limit?digit[pos]:9;
    ll ans = 0;
    for(int i=0;i<=num;i++)
    {
        if(abs(i-pre)<2)continue;
        if(lead&&i==0)
            ans += dfs(pos-1,pre,limit&&i==digit[pos],true);
        else ans+= dfs(pos-1,i,limit&&i==digit[pos],false);
    }
    if(!lead && !limit)return dp[pos][pre] = ans;
    else return ans;
}

ll solve(ll x)
{
    int len = 0;
    while(x>0)
    {
        digit[++len] = x%10;
        x/=10;
    }
    return dfs(len,-10,true,true);
}
int main(){
    ll n,m;
    memset(dp,-1,sizeof(dp));
    while(~scanf("%lld%lld",&n,&m))
    {
        printf("%lld
",solve(m)-solve(n-1));
    }
}

贴一张特判的，感觉这个目的性和思路比较清晰；

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>

using namespace std;

int a, b,num[20],dp[20][12];

int dfs(int len, int last, bool shangxian)
{
    int p;
    if (len <= 0)
        return 1;
    if (!shangxian && dp[len][last] != -1&& last >= 0)
        return dp[len][last];
    int cnt = 0, maxx = (shangxian ? num[len] : 9);
    for (int i = 0; i <= maxx; i++)
    {
        if (abs(i - last) < 2)
            continue;
        p = i;
        if (i == 0 && last == -10)
            p = last;
        cnt += dfs(len - 1, p, shangxian && (i == maxx));
    }
    //return cnt;
    if (last >= 0 && !shangxian)
        dp[len][last] = cnt;
    return cnt;
}

int solve(int x)
{
    int k = 0;
    while (x)
    {
        num[++k] = x % 10;
        x /= 10;
    }
    memset(dp, 255, sizeof(dp));
    return dfs(k, -10, true);
}

int main()
{
    scanf("%d%d", &a, &b);
    printf("%d
", solve(b) - solve(a - 1));

    return 0;
}