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Pre-knowledge: linear dp, backpack, tree dp, interval dp
Content Preview:
- Shaped pressure dp
- digital dp
- dp optimization (prefix sums, monotonic queues, slope optimization)
1. Stress dp.
Idea: If the title of the\(n\) The range is particularly small (\(<=20\)) can probably press the dp.
State: f[i][j]: considering the former\(i\) number of them, and the set that has been considered is\(j\) (j is a binary number, 1 means considered, 0 means not considered)
set of
(Supporting knowledge: bitwise operations)
Example question:
-
P1171 Salesman's Dilemma
dp[i][j] denotes the shortest path from the starting point to point j with state i at the arrival point.
Compare the simple shape of the pressure dp, look at the code between it:
Click to view code
#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) //Enumeration state
for(int j=0;j<n;j++){ //Enumerate the points reached in the next step
if(!((i >> j) & 1)) continue;
for(int k=0;k<n;k++){ Enumeration of intermediary points
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;
}
-
P1896 [SCOI2005] Non-aggression
Idea: f[i][j][s] then represents the total number of situations when only the first i rows are considered, when there are and only s kings in the first i rows (including row i), and when the situation of the king in row i is the state numbered j. And k then represents the state number of the king's situation in row i-1.
Click to view code
#include<cstdio>
#include<cstdio> #include<iostream>
#include<cstring> #include<cmath>
#include<cmath>.
#include<string>.
#include<algorithm> #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)//preprocess out each state
{
if(node>=n)//if it has been processed (note it is greater than or equal to)
{
sit[++cnt]=he;
gs[cnt]=sum.
return; // create a new state
}
dfs(he,sum,node+1);// don't use the node one
dfs(he+(1<<node),sum+1,node+2);//use the first node, add 2 to the node, and skip the next grid.
}
int main()
{
scanf("%d%d",&n,&yong);
dfs(0,0,0);
for(int i=1;i<=cnt;i++)f[1][i][gs[i]]=1;//all states in the first level are with 1 case
for(int i=2;i<=n;i++)
for(int j=1;j<=cnt;j++)
for(int k=1;k<=cnt;k++)//enumerate i, j, k
{
if(sit[j]&sit[k])continue;
if((sit[j]<<1)&sit[k])continue;
if(sit[j]&(sit[k]<<1))continue;//exclude unlawful king cases
for(int s=yong;s>=gs[j];s--)f[i][j][s]+=f[i-1][k][s-gs[j]];//enumerate s, calculate f[i][j][s]
}
long long ans=0;
for(int i=1;i<=cnt;i++)ans+=f[n][i][yong];//count final answer, remember to use long long
printf("%lld",ans);
return 0.
}
2. Digital dp.
dp with the number of digits of a number, generally for finding the number of digits in an interval that satisfies certain conditions
The interval is usually changed to\((1...r) - (1...l-1)\)
Example question:
-
P2657windy count
Recursive practices:
Click to view code
#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 = () - 1; i >= 0; i -- )
{
int x = nums[i];
for (int j = i == () - 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 < (); 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;
}
-
CF628D
Recursive approach, enumerate to that bit, now the attributes, are there any restrictions
Click to view code
#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;
}