[Template] integration plan DP Dynamic Programming
【backpack】
[01] backpack
#include<algorithm>
#include<cstdio>
int T,n,i,j,v[110],w[110],f[1010];
int main(){
scanf("%d%d",&T,&n);
for(i=1;i<=n;i++)scanf("%d%d",&v[i],&w[i]);
for(i=1;i<=n;i++)
for(j=T;j>=v[i];j--)
f[j]=std::max(f[j],f[j-v[i]]+w[i]);
printf("%d",f[T]);
}[Completely] backpack
#include<bits/stdc++.h>
using namespace std;
int T,n,ans,a[105],dp[30005];
int main(){
scanf("%d",&T);
while(T--){
memset(dp,-127,sizeof(dp));
scanf("%d",&n);dp[0]=ans=0;
for(int i=1;i<=n;i++)scanf("%d",&a[i]);
for(int i=1;i<=n;i++)
for(int j=a[i];j<=25000;j++)
dp[j]=max(dp[j],dp[j-a[i]]+1);
for(int i=1;i<=n;++i)ans+=dp[a[i]]==1;
printf("%d\n",ans);
}
}[People] backpack
Multiplayer backpack \ ([P1858] \)
#include<algorithm>
#include<cstring>
#include<cstdio>
int ans,T,n,i,j,k,K,a,b,t,re[45],v[210],w[210],f[5010][45];
int main(){
scanf("%d%d%d",&K,&T,&n);
memset(f,-127,sizeof(f));f[0][1]=0;
for(i=1;i<=n;i++)scanf("%d%d",&v[i],&w[i]);
for(i=1;i<=n;i++)
for(j=T;j>=v[i];j--){
a=b=1,t=0;
while(t<=K)
if(f[j][a]>=f[j-v[i]][b]+w[i])re[++t]=f[j][a++];
else re[++t]=f[j-v[i]][b++]+w[i];
for(k=1;k<=K;k++)f[j][k]=re[k];
}
for(i=1;i<=K;i++)ans+=f[T][i];
printf("%d",ans);
}[Packet] backpack
Jinming budget plan \ ([P1064] \)
#include<algorithm>
#include<cstdio>
using namespace std;
int T,n,i,j,a[65],s,t,r1,r2,r[65][3],v[65],w[65],f[32010];
int main(){
scanf("%d%d",&T,&n);
v[0]=0xfffffff;
for(i=1;i<=n;i++){
scanf("%d%d%d",&v[i],&s,&a[i]);
if(a[i])r[a[i]][++r[a[i]][0]]=i;w[i]=v[i]*s;
}
for(i=1;i<=n;i++)
for(j=T;j>=v[i]&&(!a[i]);j--){
t=j-v[i],r1=r[i][1],r2=r[i][2];
f[j]=max(f[j],f[t]+w[i]);
if(v[r1]<=t)f[j]=max(f[j],f[t-v[r1]]+w[i]+w[r1]);
if(v[r2]<=t)f[j]=max(f[j],f[t-v[r2]]+w[i]+w[r2]);
if(v[r1]+v[r2]<=t)f[j]=max(f[j],f[t-v[r1]-v[r2]]+w[i]+w[r1]+w[r2]);
}
printf("%d",f[T]);
}[Mixed, multiple backpack]
Because have not found mixed backpack independent, so we put them in together. . .
#include<algorithm>
#include<cstdio>
using namespace std;
int x,a,b,c,d,T,n,i,j,t,f[1010],v[100010],w[100010];
int main(){
scanf("%d:%d%d:%d%d",&a,&b,&c,&d,&n);
T=c*60+d-a*60-b;
for(i=1;i<=n;i++){
scanf("%d%d%d",&a,&b,&c);
if(!c){
for(j=0;j+a<=T;j++)
f[j+a]=max(f[j+a],f[j]+b);
continue;
}
x=1;
while(c>=x){v[++t]=x*a,w[t]=x*b,c-=x,x<<=1;}
v[++t]=c*a,w[t]=c*b;
}
for(i=1;i<=t;i++)
for(j=T;j>=v[i];j--)
f[j]=max(f[j],f[j-v[i]]+w[i]);
printf("%d",f[T]);
}[Two-dimensional] costs
Ah searched and searched to find \ (GF \) \ ([P1509] \)
#include<algorithm>
#include<cstdio>
using namespace std;
int tmp,T1,T2,x,y,n,i,j,k,v1[105],v2[105],w[105],dp[105][105],ans[105][105];
int main(){
scanf("%d",&n);
for(i=1;i<=n;i++)scanf("%d%d%d",&v1[i],&v2[i],&w[i]);
scanf("%d%d",&T1,&T2);
for(i=1;i<=n;++i)
for(j=T1;j>=v1[i];--j)
for(k=T2;k>=v2[i];--k)
if(dp[j][k]<(tmp=dp[j-v1[i]][k-v2[i]]+1)){
dp[j][k]=tmp;
ans[j][k]=ans[j-v1[i]][k-v2[i]]+w[i];
}
else if(dp[j][k]==tmp)ans[j][k]=min(ans[j][k],ans[j-v1[i]][k-v2[i]]+w[i]);
printf("%d",ans[T1][T2]);
}DP [tree]
#include<algorithm>
#include<cstring>
#include<cstdio>
#define R register int
using namespace std;
int head[6010],a,b,t,i,j,n,f[6010][2],pan[6010];
struct QAQ{int next,to;}Q[6010];
inline void add(int x,int y){Q[++t].to=y,Q[t].next=head[x],head[x]=t;}
inline void dfs(int w){
for(R k=head[w];k;k=Q[k].next){
int to=Q[k].to;
dfs(to);
f[w][0]+=max(f[to][0],f[to][1]);
f[w][1]+=f[to][0];
}
}
int main(){
scanf("%d",&n);
for(i=1;i<=n;i++)scanf("%d",&f[i][1]);
for(i=1;i<n;i++)scanf("%d%d",&a,&b),pan[a]++,add(b,a);
for(i=1;i<=n;i++)
if(!pan[i]){
dfs(i);
printf("%d\n",max(f[i][0],f[i][1]));
return 0;
}
}[Digital] DP
\(Windy\) 数 \([SCOI2009]\) \([P2657]\)
【dfs】
#include<cstring>
#include<cstdio>
#include<cmath>
#define R register int
using namespace std;
int a,b,l,num[12],dp[12][10];
inline int dfs(int len,int up,int ok){
R i,ed,ans=0;
if(len<1)return 1;
if(!ok&&~dp[len][up]&&up!=-7)return dp[len][up];
ed=ok?num[len]:9;
for(i=0;i<=ed;i++)
if(abs(i-up)>=2)
ans+=dfs(len-1,(!i&&up==-7)?up:i,ok&&(i==ed));
if(!ok&&up!=-7)dp[len][up]=ans;
return ans;
}
inline int sovle(int x){
l=0;
while(x)num[++l]=x%10,x/=10;
return dfs(l,-7,1);
}
int main(){
scanf("%d%d",&a,&b);
memset(dp,-1,sizeof(dp));
printf("%d",sovle(b)-sovle(a-1));
}【dp】
#include<cstdio>
#include<cmath>
#define R register int
using namespace std;
int i,j,k,a,b,l,ans,num[12],dp[12][10];
inline int dpp(int len){
ans=0;
for(i=1;i<len;i++)
for(j=1;j<10;j++)
ans+=dp[i][j];
for(j=1;j<num[len];j++)ans+=dp[len][j];
for(i=len-1;i>0;i--){
for(j=0;j<num[i];j++)
if(abs(num[i+1]-j)>=2)ans+=dp[i][j];
if(abs(num[i]-num[i+1])<2)break;
}
if(!i)ans++;
return ans;
}
inline int sovle(int x){
l=0;
while(x)num[++l]=x%10,x/=10;
return dpp(l);
}
inline int sakura(){
scanf("%d%d",&a,&b);
for(i=0;i<10;i++)dp[1][i]=1;
for(i=2;i<11;i++)
for(j=0;j<10;j++)
for(k=0;k<10;k++)
if(abs(j-k)>=2)
dp[i][j]+=dp[i-1][k];
printf("%d",sovle(b)-sovle(a-1));
}
int QAQWQ=sakura();
int main(){}\ (Kathy \) function \ ([HNOI2002] \) \ ([BZOJ1223] \) \ ([P2235] \)
solution to a problem: \ (Xing \) _ \ (Ling \)
[] Optimization of DP
[Dichotomy]
Longest common subsequence \ ([P1439] \)
#include<bits/stdc++.h>
using namespace std;
int pan[100010],b[100010],f[100010],n,i,a,len=1,l,r,mid;
inline int in(){
int x=0,fh=1;char c=getchar();
while(c<'0'||c>'9')c=getchar();
while(c>='0'&&c<='9')x=x*10+c-'0',c=getchar();
return x*fh;
}
int main(){
n=in();
for(i=1;i<=n;i++)a=in(),pan[a]=i,f[i]=0xfffffff;
for(i=1;i<=n;i++)a=in(),b[i]=pan[a];
f[1]=b[1];
for(i=2;i<=n;i++){
l=0;r=len;
if(b[i]>f[len])f[++len]=b[i];
else{
while(l<r){
mid=(l+r)/2;
if(f[mid]<b[i])l=mid+1;
else r=mid;
}
f[l]=min(b[i],f[l]);
}
}
printf("%d",len);
}There are like pressure \ (DP \) , the interval \ (DP \) , but the wording varied, there is no specific template amount.