Solution
T1:
設f[i][j]爲i張牌,j張黑牌的期望張數,
T2:
最難的一題(我覺得),首先這是一道DP,但最好要轉化,
令
T3:
分數規劃?vscode1183類似,二分+最短路pd,這道題目是拓撲圖,可以用類似DP的方法O(n+m)求解,注意這裏已有一個起點(我考試時多此一舉了)
CODE
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int n,m;
int main()
{
freopen("draw.in","r",stdin);
freopen("draw.out","w",stdout);
while (scanf("%d%d",&n,&m)!=EOF)
{
printf("%.4f\n",(double)m/(double)(n-m+1));
}
return 0;
}
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <cstring>
using namespace std;
const int MAXN=50;
int dp[MAXN][MAXN][MAXN][MAXN];
bool b[MAXN][MAXN];
char str[MAXN];
int n,m;
int find(int x1,int y1,int x2,int y2)
{
if (x1==x2 && y1==y2) return 0;
if (x1>x2 || y1>y2) return 0x3f3f3f3f;
if (dp[x1][y1][x2][y2]!=0x3f3f3f3f) return dp[x1][y1][x2][y2];
int ans=0x3f3f3f3f,tmp;
tmp=find(x1+1,y1,x2,y2);
for (int i=y1;i<=y2;i++)
if (b[x1][i]) tmp+=max(x2-x1,max(i-y1,y2-i));
ans=min(ans,tmp);
tmp=find(x1,y1,x2-1,y2);
for (int i=y1;i<=y2;i++)
if (b[x2][i]) tmp+=max(x2-x1,max(i-y1,y2-i));
ans=min(ans,tmp);
tmp=find(x1,y1+1,x2,y2);
for (int i=x1;i<=x2;i++)
if (b[i][y1]) tmp+=max(y2-y1,max(i-x1,x2-i));
ans=min(ans,tmp);
tmp=find(x1,y1,x2,y2-1);
for (int i=x1;i<=x2;i++)
if (b[i][y2]) tmp+=max(y2-y1,max(i-x1,x2-i));
ans=min(ans,tmp);
dp[x1][y1][x2][y2]=ans;
return ans;
}
void ReadInfo()
{
memset(dp,0x3f,sizeof(dp));
scanf("%d%d",&n,&m);
for (int i=1;i<=n;i++)
{
scanf("%s",str+1);
for (int j=1;j<=m;j++)
if (str[j]=='Y') b[i+j][i-j+m]=true;
}
}
int main()
{
freopen("flag.in","r",stdin);
freopen("flag.out","w",stdout);
ReadInfo();
cout<<find(2,1,n+m,n-1+m)<<endl;
return 0;
}
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <cstring>
using namespace std;
const double eps=1e-6;
const int MAXN=20010,MAXM=400010;
int Head[MAXN],tp[MAXN];
double dis[MAXN];
int n,m,tot,maxw,mint;
struct Edge{
int v,t,w,next;
}edge[MAXM];
void add_edge(int x,int y,int z,int w)
{
edge[++tot]=(Edge){y,z,w,Head[x]};
Head[x]=tot;
}
void ReadInfo()
{
tot=0; mint=0x3f3f3f3f; maxw=-1;
scanf("%d%d",&n,&m);
for (int i=1;i<=m;i++)
{
int x,y,z,w;
scanf("%d%d%d%d",&x,&y,&z,&w);
add_edge(x,y,z,w);
mint=min(mint,z); maxw=max(maxw,w);
}
}
bool pd(double x)
{
for (int i=1;i<=n;i++) dis[i]=-1e20;
dis[1]=0;
for (int i=1;i<=n;i++)
for (int j=Head[i];j;j=edge[j].next)
{
int v=edge[j].v,t=edge[j].t,w=edge[j].w;
double ww=w-x*t;
if (dis[v]<dis[i]+ww) dis[v]=dis[i]+ww;
}
if (dis[n]+eps>=0) return true;
else return false;
}
void solve()
{
double l,r,mid;
l=0; r=(double)maxw/(double)mint;
while (r-l>eps)
{
mid=(l+r)/2;
if (pd(mid)) l=mid;
else r=mid;
}
printf("%.4f\n",l);
}
int main()
{
freopen("treasure.in","r",stdin);
freopen("treasure.out","w",stdout);
ReadInfo();
solve();
return 0;
}