POI2012

现在才开始写 POI 是不是太弱了？

-Rendezvous

怎么说呢，我发现我的代码好长啊～长啊～长啊～长长长长长长长长长长长长长长长长长长长长长长啊～

大概就是在一个内向树上搞一个类似 lca 的东西，想想内向树估计就可以搞出来了吧……

#include <cstdio>
const int sizeOfPoint=500005;
const int sizeOfEdge=500005;

inline int lg(int);
inline int min(int, int);
inline int max(int, int);
inline void swap(int & , int & );
inline int getint();
inline void putint(int);

struct edge {int point; edge * next;};
edge memory[sizeOfEdge], * port=memory;
inline edge * newedge(int, edge * );

struct node {int x, y; inline node(int=0, int=0);};
inline bool operator < (node, node);

int n, m, k;
int p[sizeOfPoint];
edge * e[sizeOfPoint];
int a[32][sizeOfPoint];
int g[sizeOfPoint], s[sizeOfPoint];
int f[sizeOfPoint], d[sizeOfPoint];
int l[sizeOfPoint];
bool v[sizeOfPoint], b[sizeOfPoint];
inline void bfs(int);
inline int lca(int, int);

int main()
{
    n=getint(), k=getint();
    for (int i=1;i<=n;i++)
    {
        p[i]=getint();
        e[p[i]]=newedge(i, e[p[i]]);
    }

    for (int i=1;i<=n;i++) if (!v[i])
    {
        int u=i;
        for ( ;!b[u];u=p[u]) b[u]=true;

        ++m;
        for (int j=1;!l[u];j++, u=p[u])
        {
            g[u]=m;
            s[m]++;
            l[u]=j;
            bfs(u);
        }
    }

    for (int i=1;i<=k;i++)
    {
        int a=getint(), b=getint();

        if (g[f[a]]!=g[f[b]])
        {
            putint(-1), putchar(' ');
            putint(-1), putchar('
');
        }
        else if (f[a]==f[b])
        {
            int c=lca(a, b);
            putint(d[a]-d[c]), putchar(' ');
            putint(d[b]-d[c]), putchar('
');
        }
        else
        {
            int o=s[g[f[a]]];
            node ans1=node(d[a], d[b]), ans2=node(d[a], d[b]);

            if (l[f[a]]<l[f[b]])
            {
                ans1.x+=l[f[b]]-l[f[a]];
                ans2.y+=o-(l[f[b]]-l[f[a]]);
            }
            else
            {
                ans1.x+=o-(l[f[a]]-l[f[b]]);
                ans2.y+=l[f[a]]-l[f[b]];
            }

            if (ans1<ans2)
            {
                putint(ans1.x), putchar(' ');
                putint(ans1.y), putchar('
');
            }
            else
            {
                putint(ans2.x), putchar(' ');
                putint(ans2.y), putchar('
');
            }
        }
    }

    return 0;
}

inline int lg(int x)
{
    return 31-__builtin_clz(x);
}
inline int min(int x, int y)
{
    return x<y?x:y;
}
inline int max(int x, int y)
{
    return x>y?x:y;
}
inline void swap(int & x, int & y)
{
    int t=x; x=y; y=t;
}
inline int getint()
{
    register int num=0;
    register char ch;
    do ch=getchar(); while (ch<'0' || ch>'9');
    do num=num*10+ch-'0', ch=getchar(); while (ch>='0' && ch<='9');
    return num;
}
inline void putint(int num)
{
    char stack[11];
    register int top=0;
    if (num==0) stack[top=1]='0';
    if (num<0) putchar('-'), num=-num;
    for ( ;num;num/=10) stack[++top]=num%10+'0';
    for ( ;top;top--) putchar(stack[top]);
}

inline edge * newedge(int point, edge * next)
{
    edge * ret=port++;
    ret->point=point; ret->next=next;
    return ret;
}

inline node::node(int _x, int _y)
{
    x=_x;
    y=_y;
}
inline bool operator < (node a, node b)
{
    if (max(a.x, a.y)<max(b.x, b.y)) return true;
    if (max(a.x, a.y)>max(b.x, b.y)) return false;
    if (min(a.x, a.y)<min(b.x, b.y)) return true;
    if (min(a.x, a.y)>min(b.x, b.y)) return false;
    return a.y<b.y;
}

inline void bfs(int root)
{
    static int q[sizeOfPoint];
    int s=0, t=0;
    d[root]=0;
    f[root]=root;

    for (q[t++]=root;s<t;s++)
    {
        int u=q[s];
        v[u]=true;
        if (d[u]>1)
        {
            int lim=lg(d[u]);
            for (int i=1;i<=lim;i++)
                a[i][u]=a[i-1][a[i-1][u]];
        }

        for (edge * i=e[u];i;i=i->next) if (i->point!=p[u] && !l[i->point])
        {
            d[i->point]=d[u]+1;
            f[i->point]=root;
            a[0][i->point]=u;
            q[t++]=i->point;
        }
    }
}
inline int lca(int u, int v)
{
    int dist;
    if (d[u]<d[v]) swap(u, v);
    while ((dist=d[u]-d[v])) u=a[__builtin_ctz(dist)][u];
    if (u==v) return u;
    for (int i=31;i>=0;i--)
        if (a[i][u]!=a[i][v])
            u=a[i][u],
            v=a[i][v];
    return a[0][u];
}

-Distance

一道我认为做法极其诡异的题目，很容易想到若令 g[i] 表示 i 有几个质因子，则 d[x, y]=g[x]+g[y]-2*g[gcd(x, y)]

然后我就想不下去了……一看题解，居然是枚举 x 的因数！然后暴力搞搞就出来了，细节部分要仔细推敲。

还有 n 的质因子数应该是 O(log₂n) 所以 n 的因子数最差情况似乎是 O(n) 级别的？也只能说是数据弱了～

#include <cstdio>
#include <cstring>
const int sizeOfNumber=100001;
const int sizeOfSieve=1000001;
const int inf=0x7F7F7F7F;

inline int getint();
inline void putint(int);

bool b[sizeOfSieve];
int m, p[sizeOfNumber];
int e[sizeOfSieve], c[sizeOfSieve];
inline void sieve();

struct node {int v, t; inline node(int=0, int=0);};

int n, l;
int ansv, anst;
int a[sizeOfNumber];
node s[sizeOfNumber];
int v1[sizeOfSieve], v2[sizeOfSieve];
int t1[sizeOfSieve], t2[sizeOfSieve];
void search(int, int);
void query(int, int);

int main()
{
    sieve();

    n=getint();
    for (int i=1;i<=n;i++)
        a[i]=getint();

    memset(v1, 0x7F, sizeof(v1));
    memset(v2, 0x7F, sizeof(v2));
    for (int i=1;i<=n;i++)
    {
        int t=a[i];

        s[0]=node(0, 0);
        for (l=0;t>1;t/=e[t])
        {
            if (e[t]!=s[l].v)
                s[++l]=node(e[t], 1);
            else
                s[l].t++;
        }
        s[0]=node(a[i], i);
        search(1, 1);
    }
    for (int i=1;i<=n;i++)
    {
        int t=a[i];

        s[0]=node(0, 0);
        for (l=0;t>1;t/=e[t])
        {
            if (e[t]!=s[l].v || !l)
                s[++l]=node(e[t], 1);
            else
                s[l].t++;
        }
        s[0]=node(a[i], i);

        ansv=anst=inf;
        query(1, 1);
        putint(anst);
    }

    return 0;
}

inline int getint()
{
    register int num=0;
    register char ch;
    do ch=getchar(); while (ch<'0' || ch>'9');
    do num=num*10+ch-'0', ch=getchar(); while (ch>='0' && ch<='9');
    return num;
}
inline void putint(int num)
{
    char stack[11];
    register int top=0;
    if (num==0) stack[top=1]='0';
    for ( ;num;num/=10) stack[++top]=num%10+'0';
    for ( ;top;top--) putchar(stack[top]);
    putchar('
');
}

inline void sieve()
{
    for (int i=2;i<sizeOfSieve;i++)
    {
        if (!b[i])
        {
            p[m++]=i;
            e[i]=i; c[i]=1;
        }

        for (int j=0;j<m;j++)
        {
            if (i*p[j]>=sizeOfSieve) break;
            b[i*p[j]]=true;
            e[i*p[j]]=p[j]; c[i*p[j]]=c[i]+1;
            if (e[i]==p[j]) break;
        }
    }
}

inline node::node(int _v, int _t)
{
    v=_v, t=_t;
}

void search(int v, int t)
{
    if (t>l)
    {
        int g=c[s[0].v]-(c[v]<<1);
        if (g<v1[v])
        {
            v2[v]=v1[v], t2[v]=t1[v];
            v1[v]=g, t1[v]=s[0].t;
        }
        else if (g<v2[v])
            v2[v]=g, t2[v]=s[0].t;
    }
    else
    {
        for (int i=0, j=1;i<=s[t].t;i++, j=j*s[t].v)
            search(v*j, t+1);
    }
}
void query(int v, int t)
{
    if (t>l)
    {
        if (s[0].t!=t1[v])
        {
            if (v1[v]<ansv || (v1[v]==ansv && t1[v]<anst))
                ansv=v1[v], anst=t1[v];
        }
        else if (v2[v]<ansv || (v2[v]==ansv && t2[v]<anst))
                ansv=v2[v], anst=t2[v];
    }
    else
    {
        for (int i=0, j=1;i<=s[t].t;i++, j=j*s[t].v)
            query(v*j, t+1);
    }
}

-Letters

这道题目应该还是比较简单的吧，一眼就看出是求逆序对数，利用第一个字符串给第二个字符串标号（是字母对应字母的标号，开 26 个队列即可）

本质其实和直接算逆序对数是一样的，只不过这道题可以看作是对 “顺序” 重新下了一个定义罢了～

#include <cstdio>
#include <cstring>
#include <queue>
typedef long long LL;
const int sizeOfString=1000001;

inline int getint();
inline int getstr(char * );
inline void putint(LL);

int C[sizeOfString];
inline int lowbit(int);
inline void update(int);
inline int query(int);

int N;
char J[sizeOfString], M[sizeOfString];
std::queue<int> q[26];
int A[sizeOfString];

int main()
{
    LL ans=0;

    N=getint();
    getstr(J);
    getstr(M);

    for (int i=0;i<N;i++)
        q[J[i]-'A'].push(i);
    for (int i=0;i<N;i++)
    {
        A[i]=1+q[M[i]-'A'].front();
        q[M[i]-'A'].pop();
    }

    for (int i=0;i<N;i++)
    {
        ans+=query(A[i]);
        update(A[i]);
    }

    putint(ans);

    return 0;
}

inline int getint()
{
    register int num=0;
    register char ch;
    do ch=getchar(); while (ch<'0' || ch>'9');
    do num=num*10+ch-'0', ch=getchar(); while (ch>='0' && ch<='9');
    return num;
}
inline int getstr(char * str)
{
    register int len=0;
    register char ch;
    do ch=getchar(); while (ch<'A' || ch>'Z');
    do str[len++]=ch, ch=getchar(); while (ch>='A' && ch<='Z');
    return len;
}
inline void putint(LL num)
{
    char stack[22];
    register int top=0;
    if (num==0) stack[top=1]='0';
    for ( ;num;num/=10) stack[++top]=num%10+'0';
    for ( ;top;top--) putchar(stack[top]);
    putchar('
');
}

inline int lowbit(int i)
{
    return i & -i;
}
inline void update(int i)
{
    for ( ;i;i-=lowbit(i))
        C[i]++;
}
inline int query(int i)
{
    int ret=0;
    for ( ;i<=N;i+=lowbit(i))
        ret+=C[i];
    return ret;
}