【NOIP2017】Day6

Solution

T1：
设f[i][j]为i张牌，j张黑牌的期望张数，

f[i][j]==i/j∗f[i−1][j−1]

数学归纳法可以证明f[n][m]=m/(n-m+1)。
T2：
最难的一题（我觉得），首先这是一道DP，但最好要转化，
令x′=x+y y′=x−y 这样曼哈顿距离就变成了max{|x1−x2|,|y1−y2|}，这是我们对新图DP（多出来的不要紧，全部一起做了），DP四维表示一个矩阵，推的时候每一行的点取到另外三边的最大值的最小值，考虑没包全的情况虽然会偏大，但正确ans会出现，所以错误的会没有，所以这个算法是对的

T3：
分数规划？vscode1183类似，二分+最短路pd，这道题目是拓扑图，可以用类似DP的方法O（n+m)求解，注意这里已有一个起点（我考试时多此一举了）

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;}