A preliminary study on pressure DP

Wrote a few lines of pressure ... And then it's been decadent ...

2064: Split

http://www.lydsy.com/JudgeOnline/problem.php?id=2064

The initial is positive, the last is negative, assuming we can find the K group to make up 0, the answer is n+m-2*k. So the pressure of the shape. I don't really have a thing.

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define rep(i,l,r) for (int i=l;i<=r;i++) #define maxn 1<<22 using namespace std; int t,mm,f[maxn],sum[maxn],ans,n,m,a[50],b[50]; int main(){      scanf ( "%d" ,&n);      rep(i,0,n-1) scanf ( "%d" ,&a[i]);      scanf ( "%d" ,&m);      rep(i,0,m-1) scanf ( "%d" ,&b[i]);      rep(i,0,n-1) sum[1<<i]=a[i];      rep(i,n,n+m-1) sum[1<<i]=-b[i-n];      mm=1<<(n+m);      rep(i,1,mm-1){          t=i&(-i);          sum[i]=sum[t]+sum[i-t];          rep(j,0,m+n-1)              if ((1<<j)&i) f[i]=max(f[i],f[i^(1<<j)]);          if (!sum[i]) ++f[i];         }      printf ( "%d\n" ,n+m-2*f[mm-1]);      return 0; }1725: [Usaco2006 nov]corn Field Ranch arrangement Http://www.lydsy.com/judgeonline/problem.php?id=1725f[i][j] denotes line I, the case is J's answer. So bare-shaped pressure. #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define rep(i,l,r) for (int i=l;i<=r;i++) #define maxn 5000 #define mm 100000000 int y,sum,tot,ans,ff[20][maxn],f[20][maxn],g[20][maxn],d[maxn],map[20][20],n,m; using namespace std; void dfs( int x){      if (x>m) {          d[y]=++tot;          g[y][tot]=sum;          ff[y][tot]=ans;          return ;      }      if (map[y][x]==1){          sum+=(1<<(x-1));          ans++;          dfs(x+2);          sum-=(1<<(x-1));          ans--;      }      dfs(x+1); } int main(){      scanf ( "%d%d" ,&n,&m);      rep(i,1,n)          rep(j,1,m) scanf ( "%d" ,&map[i][j]);      rep(i,1,n){          tot=ans=sum=0;          y=i;          dfs(1);      }      ans=0;      if (n==1) printf ( "%d\n" ,d[1]);      else {          rep(i,1,d[1]) f[1][i]=1;          rep(i,2,n)              rep(j,1,d[i])                  rep(k,1,d[i-1]){                      if ((g[i][j]&g[i-1][k])==0) f[i][j]=(f[i][j]+f[i-1][k])%mm;                  }            rep(i,1,d[n]) ans=(ans+f[n][i])%mm;          printf ( "%d" ,ans);      }      return 0; }1231: [Usaco2008 nov]mixup2 chaotic Cow http://www.lydsy.com/JudgeOnline/problem.php?id=1231 #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<cmath> #define rep(i,l,r) for (int i=l;i<=r;i++) #define ll long long #define maxn 70000 ll f[maxn][20],mm,ans; int n,m,a[20]; int main(){      scanf ( "%d%d" ,&n,&m);      rep(i,1,n) scanf ( "%d" ,&a[i]);      rep(i,0,n-1) f[1<<i][i+1]=1;      mm=(1<<n)-1;      rep(i,0,mm)          rep(j,1,n) if (1<<(j-1)&i)              rep(k,1,n) if ((1<<(k-1)|i)!=i&& abs (a[k]-a[j])>m) f[i|1<<(k-1)][k]+=f[i][j];      rep(i,1,n) ans+=f[mm][i];        printf ( "%lld\n" ,ans);      return 0; }2073: [poi2004]przhttp://www.lydsy.com/judgeonline/problem.php?id=2073 is more magical when the enumeration subset is transferred. Actually, it's nothing. #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<cmath> #define rep(i,l,r) for (int i=l;i<=r;i++) #define ll long long #define maxn 70000 using namespace std; int mm,n,m,a[20],b[20],t[20],w[maxn],tim[maxn],f[maxn]; int main(){      scanf ( "%d%d" ,&m,&n);      rep(i,1,n) scanf ( "%d%d" ,&a[i],&b[i]),t[i]=1<<(i-1);      mm=(1<<n)-1;      rep(i,0,mm)          rep(j,1,n) if (t[j]&i) tim[i]=max(tim[i],a[j]),w[i]+=b[j];      memset (f,127/3, sizeof (f)); f[0]=0;      rep(i,1,mm)          for ( int j=i;j;j=i&(j-1))              if (w[j]<=m) f[i]=min(f[i],tim[j]+f[i^j]);      printf ( "%d\n" ,f[mm]);      return 0; }

A preliminary study on pressure DP