CF 105 div 2 (dp水題）

寫過史上最簡單的一套cf

代碼量少，想也好想

D

概率dp，轉移只用考慮公主就可以了

#include<iostream>
#define N 1005
using namespace std;
double dp[N][N];
bool vis[N][N];
double dfs(int nw, int nb){
	if (vis[nw][nb]) return dp[nw][nb];
	if (nw==0) return 0;
	if (nb==0) return 1;
	double w=nw, b=nb,ans=0;
	if (nw) ans+=(w/(w+b)); //直接抽到w 
	if (nb>=3) ans+=(b/(w+b))*((b-1)/(w+b-1))*((b-2)/(w+b-2))*dfs(nw,nb-3); //抽到b 
	if (nw>=1&&nb>=2) ans+=(b/(w+b))*((b-1)/(w+b-1))*(w/(w+b-2))*dfs(nw-1,nb-2);
	vis[nw][nb]=true;
	return dp[nw][nb]=ans;
}
int main(){
	int w,b;
	cin>>w>>b;
	dp[w][b]=dfs(w,b);
	printf("%.9lf",dp[w][b]);
}

E

先預處理每行，然後簡單dp

#include<iostream>
#define N 105
#define M 10005
using namespace std;

int dp[N][N],a[N],h[N],l[N],now[N];
int f[N][M];
int main(){
	int n,m;
	cin>>n>>m;
	for (int i=1;i<=n;i++){
		cin>>now[i];
		for (int j=1;j<=now[i];j++){
			cin>>a[j];
			h[j]=h[j-1]+a[j];
		}
		l[now[i]+1]=0;
		for (int j=now[i];j>=1;j--){
			l[j]=l[j+1]+a[j];
		}
		dp[i][now[i]]=h[now[i]];
		for (int j=1;j<now[i];j++){
			for (int k=0;k<=j;k++){
				dp[i][j]=max(dp[i][j],h[k]+l[now[i]-(j-k)+1]);
			}
		}
	}
	int tot=0;
	for (int i=1;i<=n;i++){
		tot+=now[i];
		for (int j=0;j<=min(tot,m);j++){
			for (int k=0;k<=min(j,now[i]);k++){
				f[i][j]=max(f[i][j],f[i-1][j-k]+dp[i][k]);
			}
		}
	}
	cout<<f[n][m]<<endl;
}