3月部分题目简要题解

hdu5820

官方题解：http://www.cnblogs.com/duoxiao/p/5777700.html

#include<bits/stdc++.h>

using namespace std;
struct node{int l,r,s;} tr[500010*40];
vector<int> g[50010];
int pre[50010],h[50010],n,t;
int build(int l,int r)
{
    ++t; tr[t].s=0;
    if (l==r) return t;
    int m=(l+r)>>1,q=t;
    tr[t].l=build(l,m);
    tr[t].r=build(m+1,r);
    return q;
}

int ask(int i,int l,int r,int x,int y)
{
    if (x>y) return 0;
    if (x<=l&&y>=r) return tr[i].s;
    else {
        int m=(l+r)>>1,s=0;
        if (x<=m) s+=ask(tr[i].l,l,m,x,y);
        if (y>m) s+=ask(tr[i].r,m+1,r,x,y);
        return s;
    }
}

int add(int i,int l,int r,int x)
{
    ++t;
    if (l==r)
    {
        tr[t].s=tr[i].s+1;
        return t;
    }
    int m=(l+r)>>1,q=t;
    if (x<=m)
    {
        tr[q].r=tr[i].r;
        tr[q].l=add(tr[i].l,l,m,x);
    }
    else {
        tr[q].l=tr[i].l;
        tr[q].r=add(tr[i].r,m+1,r,x);
    }
    tr[q].s=tr[tr[q].l].s+tr[tr[q].r].s;
    return q;
}


bool check()
{
    memset(pre,0,sizeof(pre));
    for (int i=1; i<=50000; i++)
    {
        h[i]=h[i-1];
        for (int j=0; j<g[i].size(); j++)
        {
            int l=(!j)?0:g[i][j-1];
            int r=(j==g[i].size()-1)?50001:g[i][j+1];
           // cout <<l<<" "<<r<<" "<<pre[g[i][j]]<<" "<<ask(h[i],1,50000,l+1,r-1)<<endl;
            if (ask(h[i],1,50000,l+1,r-1)!=ask(h[pre[g[i][j]]],1,50000,l+1,r-1)) return 0;
        }
        for (int j=0; j<g[i].size(); j++)
        {
            h[i]=add(h[i],1,50000,g[i][j]);
            pre[g[i][j]]=i;
        }
    }
    return 1;
}

int main()
{
    scanf("%d",&n);
    t=0;
    h[0]=build(1,50000); int tmp=t;
    while (n)
    {
        t=tmp;
        for (int i=1; i<=n; i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            g[x].push_back(y);
        }
        for (int i=1; i<=50000; i++)
        {
            sort(g[i].begin(),g[i].end());
            g[i].erase(unique(g[i].begin(),g[i].end()),g[i].end());
        }
        if (check()) puts("YES"); else puts("NO");
        for (int i=1; i<=50000; i++) g[i].clear();
        scanf("%d",&n);
    }
}

hdu5788

官方题解：http://www.cnblogs.com/duoxiao/p/5777661.html

其实有个简单的方法求子树的[t/2]名次：直接在dfs序上建立可持久化线段树查询即可

#include<bits/stdc++.h>

using namespace std;
typedef long long ll;
struct way{int po,next;} e[100010];
struct node{int s,l,r;} tr[100010*40];
int p[100010],l[100010],h[100010],w[100010],cur[100010],a[100010],r[100010],len,n,tot,t;
ll c[100010],ans;
void add(int x,int y)
{
    e[++len].po=y;
    e[len].next=p[x];
    p[x]=len;
}

int build(int l,int r)
{
    ++t; tr[t].s=0;
    if (l==r) return t;
    int m=(l+r)>>1,q=t;
    tr[t].l=build(l,m);
    tr[t].r=build(m+1,r);
    return q;
}

int mid(int a,int b,int l,int r,int k)
{
    if (l==r) return l;
    else {
        int m=(l+r)>>1,s=tr[tr[b].l].s-tr[tr[a].l].s;
        if (k<=s) return mid(tr[a].l,tr[b].l,l,m,k);
        else return mid(tr[a].r,tr[b].r,m+1,r,k-s);
    }
}

int add(int i,int l,int r,int x)
{
    ++t;
    if (l==r)
    {
        tr[t].s=tr[i].s+1;
        return t;
    }
    int m=(l+r)>>1,q=t;
    if (x<=m)
    {
        tr[q].r=tr[i].r;
        tr[q].l=add(tr[i].l,l,m,x);
    }
    else {
        tr[q].l=tr[i].l;
        tr[q].r=add(tr[i].r,m+1,r,x);
    }
    tr[q].s=tr[tr[q].l].s+tr[tr[q].r].s;
    return q;
}

void dfs1(int x)
{
    l[x]=++tot; h[tot]=add(h[tot-1],1,100000,a[x]);
    for (int i=p[x]; i; i=e[i].next)
    {
        int y=e[i].po;
        dfs1(y);
    }
    w[x]=mid(h[l[x]-1],h[tot],1,100000,(tot-l[x]+2)/2);
    if (l[x]==tot) cur[x]=0; else cur[x]=mid(h[l[x]-1],h[tot],1,100000,(tot-l[x]+2)/2+1)-w[x];
}

void work(int x,int w)
{
    for (int i=x; i; i-=i&(-i)) c[i]+=w;
}

ll ask(int x)
{
    ll s=0;
    for (int i=x; i<=100000; i+=i&(-i)) s+=c[i];
    return s;
}

void dfs2(int x)
{
    ll wh=(p[x]==0)?100000-a[x]:cur[x]-w[x];
    ans=max(ans,ask(a[x])+wh);
    if (cur[x]) work(w[x],cur[x]);
    for (int i=p[x]; i; i=e[i].next)
    {
        int y=e[i].po;
        dfs2(y);
    }
    if (cur[x]) work(w[x],-cur[x]);
}

int main()
{
    t=0; h[0]=build(1,100000);
    int tmp=t;
    while (scanf("%d",&n)!=EOF)
    {
        t=tmp;
        len=0; memset(p,0,sizeof(p));
        for (int i=1; i<=n; i++) scanf("%d",&a[i]);
        for (int i=2; i<=n; i++)
        {
            int fa;
            scanf("%d",&fa);
            add(fa,i);
        }
        tot=0; h[1]=h[0]; dfs1(1);
        ll s=0;
        for (int i=1; i<=n; i++) s+=w[i];
     //   for (int i=1; i<=n; i++) cout <<"*"<<w[i]<<" "<<cur[i]<<endl;
        memset(c,0,sizeof(c)); ans=0;
        dfs2(1);
        printf("%lld
",ans+s);
    }
}

hdu3605

把n按照可选择方案二进制状态压缩，建图最大流即可

#include<bits/stdc++.h>

using namespace std;
const int inf=1e9+7;
struct way{int po,flow,next;} e[2400010];
int t,n,p[2010],pre[2010],numh[2010],cur[2010],h[2010],d[2010],b[2010],m,len;

void add(int x,int y,int f)
{
     e[++len].po=y;
     e[len].flow=f;
     e[len].next=p[x];
     p[x]=len;
}
void build(int x, int y, int f)
{
     add(x,y,f);
     add(y,x,0);
}
int sap()
{
    memset(numh,0,sizeof(numh));
    memset(h,0,sizeof(h));
    numh[0]=t+1;
    for (int i=0; i<=t; i++) cur[i]=p[i];
    int j,u=0,s=0,neck=inf;
    while (h[0]<t+1)
    {
          d[u]=neck;
          bool ch=1;
          for (int i=cur[u]; i!=-1; i=e[i].next)
          {
              j=e[i].po;
              if (e[i].flow>0&&h[u]==h[j]+1)
              {
                 neck=min(neck,e[i].flow);
                 cur[u]=i;
                 pre[j]=u;  u=j;
                 if (u==t)
                 {
                    s+=neck;
                    while (u)
                    {
                          u=pre[u];
                          j=cur[u];
                          e[j].flow-=neck;
                          e[j^1].flow+=neck;
                    }
                    neck=inf;
                 }
                 ch=0;
                 break;
              }
          }
          if (ch)
          {
             if (--numh[h[u]]==0) return s;
             int q=-1,tmp=t;
             for (int i=p[u]; i!=-1; i=e[i].next)
             {
                 j=e[i].po;
                 if (e[i].flow&&h[j]<tmp)
                 {
                    tmp=h[j];
                    q=i;
                 }
             }
             cur[u]=q; h[u]=tmp+1;
             numh[h[u]]++;
             if (u)
             {
                u=pre[u];
                neck=d[u];
             }
          }
    }
    return s;
}

int main()
{
    while (scanf("%d%d",&n,&m)!=EOF)
    {
        len=-1;
        memset(p,255,sizeof(p));
        memset(b,0,sizeof(b));
        for (int i=1; i<=n; i++)
        {
            int st=0,x;
            for (int j=0; j<m; j++)
            {
                scanf("%d",&x);
                st|=(x&1)<<j;
            }
            b[st]++;
        }
        t=(1<<m)+m+1;
        for (int i=0; i<(1<<m); i++)
        {
            for (int j=0; j<m; j++)
                if ((i>>j)&1) build(i+m+1,j+1,b[i]);
            build(0,i+m+1,b[i]);
        }
        for (int i=1; i<=m; i++)
        {
            int x;
            scanf("%d",&x);
            build(i,t,x);
        }
        if (sap()==n) puts("YES"); else puts("NO");
    }
}

hdu5988

费用取对数，简单的费用流

#include<bits/stdc++.h>

using namespace std;

using namespace std;
struct way{int po,next,flow; double cost;} e[200010];
const int inf=100000007;
const double eps=1e-7;
int pre[110],p[110],cur[110],q[4000010];
double d[110];
bool v[110];
int len,n,m,t,k;

void add(int x,int y,int f,double c)
{
     e[++len].po=y;
     e[len].flow=f;
     e[len].cost=c;
     e[len].next=p[x];
     p[x]=len;
}

void build(int x,int y, int f, double c)
{
     add(x,y,f,c);
     add(y,x,0,-c);
}

bool spfa()
{
     int f=1,r=1;
     for (int i=1; i<=t; i++) d[i]=inf;
     memset(v,0,sizeof(v));
     d[0]=0; q[1]=0;
     while (f<=r)
     {
           int x=q[f++];
           v[x]=0;
           for (int i=p[x]; i!=-1; i=e[i].next)
           {
               int y=e[i].po;
               if (e[i].flow&&d[x]+e[i].cost+eps<d[y])
               {
                  d[y]=d[x]+e[i].cost;
                  pre[y]=x; cur[y]=i;
                  if (!v[y])
                  {
                     q[++r]=y;
                     v[y]=1;
                  }
               }
           }
     }
     return d[t]<inf;
}

double mincost()
{
    int j;
    double s=0;
    while (spfa())
    {
          int neck=inf;
          for (int i=t; i; i=pre[i])
          {
              j=cur[i];
              neck=min(neck,e[j].flow);
          }
          s+=d[t]*neck;
          for (int i=t; i; i=pre[i])
          {
              j=cur[i];
              e[j].flow-=neck;
              e[j^1].flow+=neck;
          }
    }
    return s;
}

int main()
{
    int cas;
    scanf("%d",&cas);
    while (cas--)
    {
        scanf("%d%d",&n,&m);
        t=n+1; len=-1; memset(p,255,sizeof(p));
        for (int i=1; i<=n; i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            if (x-y>0) build(0,i,x-y,0);
            else if (x-y<0) build(i,t,y-x,0);
        }
        for (int i=1; i<=m; i++)
        {
            int x,y,f; double c;
            scanf("%d%d%d%lf",&x,&y,&f,&c);
            c=-log2(1.0-c);
            if (f) build(x,y,1,0);
            if (f>1) build(x,y,f-1,c);
        }
        double ans=mincost();
        ans=pow(2,-ans);
        printf("%.2lf
",1.0-ans);
    }
}

hdu5985

无穷项一般要么高斯消元要么取有限项保留对应精度即可，这里只要求出h[i][k]表示第i个在第k次操作后被全部移走的概率就可以轻易计算出答案

#include<bits/stdc++.h>

using namespace std;
double h[15][80],ans[15],p[15];
int a[15],n;
int main()
{
    int cas;
    scanf("%d",&cas);
    while (cas--)
    {
        scanf("%d",&n);
        for (int i=1; i<=n; i++)
            scanf("%d%lf",&a[i],&p[i]);
        if (n==1)
        {
            printf("%.6lf
",1.0);
            continue;
        }
        for (int i=1; i<=n; i++)
        {
            double tmp=1.0;
            for (int j=1; j<=76; j++)
            {
                tmp*=p[i];
                h[i][j]=pow(1-tmp,a[i]);
            }
        }
        for (int i=1; i<=n; i++)
        {
            ans[i]=0;
            for (int k=1; k<=75; k++)
            {
                double tmp=1.0;
                for (int j=1; j<=n; j++)
                    if (j!=i) tmp*=h[j][k];
                ans[i]+=(h[i][k+1]-h[i][k])*tmp;
            }
        }
        for (int i=1; i<=n; i++)
        {
            printf("%.6lf",ans[i]);
            if (i==n) puts(""); else printf(" ");
        }
    }
}