DP进阶合集

#include<bits/stdc++.h>
using namespace std;
int f[1<<20][20],w[20][20],n;
int main(){
	cin>>n;	
	memset(f,0x3f,sizeof f);
	f[1][0]=0;
	for(int i=0;i<n;++i)
		for(int j=0;j<n;++j)
			cin>>w[i][j];
	for(int i=1;i<(1<<n);i+=2)   //枚举状态
		for(int j=0;j<n;j++){    //枚举下一步到达的点
			if(!((i >> j) & 1)) continue; 
			for(int k=0;k<n;k++){    枚举中介点
				if(j==k) continue;
				if(!(i>>k &1)) continue; 
				f[i][j]=min(f[i][j],f[i^(1<<j)][k]+w[k][j]);
			}
		}
	int minn=2e9;
	for(int i=0;i<=n-1;++i) minn=min(minn,f[(1<<n)-1][i]+w[i][0]);
	cout<<minn<<endl;
	return 0;
}

	#include<cstdio>
	#include<iostream>
	#include<cstring>
	#include<cmath>
	#include<string>
	#include<algorithm>
	using namespace std;
	int sit[2000],gs[2000];
	int cnt=0;
	int n,yong;
	long long f[10][2000][100]={0};
	void dfs(int he,int sum,int node)//预处理出每一个状态
	{
	if(node>=n)//如果已经处理完毕（注意是大于等于）
	{
	sit[++cnt]=he;
	gs[cnt]=sum;
	return;//新建一个状态
	}
	dfs(he,sum,node+1);//不用第node个
	dfs(he+(1<<node),sum+1,node+2);//用第node个，此时node要加2，及跳过下一个格子
	}
	int main()
	{
	scanf("%d%d",&n,&yong);
	dfs(0,0,0);
	for(int i=1;i<=cnt;i++)f[1][i][gs[i]]=1;//第一层的所有状态均是有1种情况的
										 for(int i=2;i<=n;i++)
	for(int j=1;j<=cnt;j++)
						   for(int k=1;k<=cnt;k++)//枚举i、j、k
	{
	if(sit[j]&sit[k])continue;
	if((sit[j]<<1)&sit[k])continue;
	if(sit[j]&(sit[k]<<1))continue;//排除不合法国王情况
	for(int s=yong;s>=gs[j];s--)f[i][j][s]+=f[i-1][k][s-gs[j]];//枚举s，计算f[i][j][s]
	}
	long long ans=0;
	for(int i=1;i<=cnt;i++)ans+=f[n][i][yong];//统计最终答案，记得用long long
							   printf("%lld",ans);
							   return 0;
}

#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>

using namespace std;

const int N = 11;

int f[N][10];

void init()
{
    for (int i = 0; i <= 9; i ++ ) f[1][i] = 1;

    for (int i = 2; i < N; i ++ )
        for (int j = 0; j <= 9; j ++ )
            for (int k = 0; k <= 9; k ++ )
                if (abs(j - k) >= 2)
                    f[i][j] += f[i - 1][k];
}

int dp(int n)
{
    if (!n) return 0;

    vector<int> nums;
    while (n) nums.push_back(n % 10), n /= 10;

    int res = 0;
    int last = -2;
    for (int i = nums.size() - 1; i >= 0; i -- )
    {
        int x = nums[i];
        for (int j = i == nums.size() - 1; j < x; j ++ )
            if (abs(j - last) >= 2)
                res += f[i + 1][j];

        if (abs(x - last) >= 2) last = x;
        else break;

        if (!i) res ++ ;
    }


    for (int i = 1; i < nums.size(); i ++ )
        for (int j = 1; j <= 9; j ++ )
            res += f[i][j];

    return res;
}

int main()
{
    init();

    int l, r;
    cin >> l >> r;
    cout << dp(r) - dp(l - 1) << endl;

    return 0;
}

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int mod = 1e9+7;
ll m,d,a[2005],len,f[2005][2005];
char l[2005],r[2005];
ll dfs(int u,int sum,int yaoqiu){
	if(u>len) return sum==0?1:0;
	if(!yaoqiu&&f[u][sum]!=-1) return f[u][sum];
	ll tmp=0,maxx=yaoqiu==1?a[u]:9;
	if(u%2==1){
		for(int i=0;i<=maxx;i++)
			if(i!=d) tmp=(tmp+dfs(u+1,(sum*10+i)%m,yaoqiu&&(i==maxx)))%mod;
	}
	else{
		for(int i=0;i<=maxx;i++)
			if(i==d) tmp=(tmp+dfs(u+1,(sum*10+i)%m,yaoqiu&&(i==maxx)))%mod;
	}
	if(!yaoqiu) f[u][sum]=tmp;
	return tmp;
}
ll dp(char *s){
	memset(f,-1,sizeof(f)); 
	len=strlen(s+1);
	for(int i=1;i<=strlen(s+1);i++) a[i]=s[i]-'0';
	return dfs(1,0,1);
}
bool check(char *s){
    len=strlen(s+1);
    int x=0;
    for(int i=1;i<=len;i++){
        int y=s[i]-'0';
        x=(x*10+y)%m;
        if(i&1){
            if(y==d)
                return false;
        }
        else{
            if(y!=d)
                return false;
        }
    }
    return !x;
}
int main(){
	cin>>m>>d>>l+1>>r+1;
	cout<<(dp(r)-dp(l)+check(l)+mod)%mod;
    return 0;
}