ASC 16

A Cactus

给一棵树，可以连边（不能连重边），问连出仙人掌的方案数。
把模型转化为，给树染色，要求相同颜色的形成一条长度大于1的链或单点，树形dp

#include<bits/stdc++.h>using namespace std;const double eps=1e-10;const double pi=3.1415926535897932384626433832795;const double eln=2.718281828459045235360287471352;#define IN freopen("cactus.in", "r", stdin)#define OUT freopen("cactus.out", "w", stdout)#define scan(x) scanf("%d", &x)#define mp make_pair#define pb push_back#define sqr(x) (x) * (x)#define pr(x) printf("Case %d: ",x)#define prn(x) printf("Case %d:\n",x)#define prr(x) printf("Case #%d: ",x)#define prrn(x) printf("Case #%d:\n",x)#define lowbit(x) (x&(-x))#define fi first#define se secondtypedef long long ll;typedef pair<int,int> pii;typedef vector<int> vi;struct BigInteger {    typedef unsigned long long LL;    static const int BASE = 100000000;    static const int WIDTH = 8;    vector<int> s;    BigInteger& clean(){while(!s.back()&&s.size()>1)s.pop_back(); return *this;}    BigInteger(LL num = 0) {*this = num;}    BigInteger(string s) {*this = s;}    BigInteger& operator = (long long num) {        s.clear();        do {            s.push_back(num % BASE);            num /= BASE;        } while (num > 0);        return *this;    }    BigInteger& operator = (const string& str) {        s.clear();        int x, len = (str.length() - 1) / WIDTH + 1;        for (int i = 0; i < len; i++) {            int end = str.length() - i*WIDTH;            int start = max(0, end - WIDTH);            sscanf(str.substr(start,end-start).c_str(), "%d", &x);            s.push_back(x);        }        return (*this).clean();    }    BigInteger operator + (const BigInteger& b) const {        BigInteger c; c.s.clear();        for (int i = 0, g = 0; ; i++) {            if (g == 0 && i >= s.size() && i >= b.s.size()) break;            int x = g;            if (i < s.size()) x += s[i];            if (i < b.s.size()) x += b.s[i];            c.s.push_back(x % BASE);            g = x / BASE;        }        return c;    }    BigInteger operator - (const BigInteger& b) const {        assert(b <= *this); // 减数不能大于被减数        BigInteger c; c.s.clear();        for (int i = 0, g = 0; ; i++) {            if (g == 0 && i >= s.size() && i >= b.s.size()) break;            int x = s[i] + g;            if (i < b.s.size()) x -= b.s[i];            if (x < 0) {g = -1; x += BASE;} else g = 0;            c.s.push_back(x);        }        return c.clean();    }    BigInteger operator * (const BigInteger& b) const {        int i, j; LL g;        vector<LL> v(s.size()+b.s.size(), 0);        BigInteger c; c.s.clear();        for(i=0;i<s.size();i++) for(j=0;j<b.s.size();j++) v[i+j]+=LL(s[i])*b.s[j];        for (i = 0, g = 0; ; i++) {            if (g ==0 && i >= v.size()) break;            LL x = v[i] + g;            c.s.push_back(x % BASE);            g = x / BASE;        }        return c.clean();    }    BigInteger operator / (const BigInteger& b) const {        assert(b > 0);  // 除数必须大于0        BigInteger c = *this;       // 商:主要是让c.s和(*this).s的vector一样大        BigInteger m;               // 余数:初始化为0        for (int i = s.size()-1; i >= 0; i--) {            m = m*BASE + s[i];            c.s[i] = bsearch(b, m);            m -= b*c.s[i];        }        return c.clean();    }    BigInteger operator / (const ll& b) const {        assert(b > 0);  // 除数必须大于0        BigInteger c = *this;       // 商:主要是让c.s和(*this).s的vector一样大        ll m=0;               // 余数:初始化为0        for (int i = s.size()-1; i >= 0; i--) {            m = m*BASE + s[i];            c.s[i] = m/b ;            m -= b*c.s[i];        }        return c.clean();    }    BigInteger operator % (const ll& b) const {        assert(b > 0);  // 除数必须大于0        BigInteger c = *this;       // 商:主要是让c.s和(*this).s的vector一样大        ll m=0;                       for (int i = s.size()-1; i >= 0; i--) {            m = m*BASE + s[i];            m%=b;        }        return BigInteger(m);    }    BigInteger operator % (const BigInteger& b) const { //方法与除法相同        BigInteger c = *this;        BigInteger m;        for (int i = s.size()-1; i >= 0; i--) {            m = m*BASE + s[i];            c.s[i] = bsearch(b, m);            m -= b*c.s[i];        }        return m;    }    // 二分法找出满足bx<=m的最大的x    int bsearch(const BigInteger& b, const BigInteger& m) const{        int L = 0, R = BASE-1, x;        while (1) {            x = (L+R)>>1;            if (b*x<=m) {if (b*(x+1)>m) return x; else L = x;}            else R = x;        }    }    BigInteger& operator += (const BigInteger& b) {*this = *this + b; return *this;}    BigInteger& operator -= (const BigInteger& b) {*this = *this - b; return *this;}    BigInteger& operator *= (const BigInteger& b) {*this = *this * b; return *this;}    BigInteger& operator /= (const BigInteger& b) {*this = *this / b; return *this;}    BigInteger& operator %= (const BigInteger& b) {*this = *this % b; return *this;}    BigInteger& operator /= (const ll& b) {*this = *this / b; return *this;}    BigInteger& operator %= (const ll& b) {*this = *this % b; return *this;}    bool operator < (const BigInteger& b) const {        if (s.size() != b.s.size()) return s.size() < b.s.size();        for (int i = s.size()-1; i >= 0; i--)            if (s[i] != b.s[i]) return s[i] < b.s[i];        return false;    }    bool operator >(const BigInteger& b) const{return b < *this;}    bool operator<=(const BigInteger& b) const{return !(b < *this);}    bool operator>=(const BigInteger& b) const{return !(*this < b);}    bool operator!=(const BigInteger& b) const{return b < *this || *this < b;}    bool operator==(const BigInteger& b) const{return !(b < *this) && !(b > *this);}};ostream& operator << (ostream& out, const BigInteger& x) {    out << x.s.back();    for (int i = x.s.size()-2; i >= 0; i--) {        char buf[20];        sprintf(buf, "%08d", x.s[i]);        for (int j = 0; j < strlen(buf); j++) out << buf[j];    }    return out;}istream& operator >> (istream& in, BigInteger& x) {    string s;    if (!(in >> s)) return in;    x = s;    return in;}const int maxn=205;BigInteger dp[maxn][4];vi g[maxn];int n;void DFS(int u,int f){    if(g[u].size()==1 && g[u][0]==f)    {        dp[u][1]=0;        dp[u][0]=0;        dp[u][2]=1;        dp[u][3]=0;        return;    }    dp[u][2]=1;dp[u][0]=dp[u][1]=dp[u][3]=0;    for(int p : g[u])        if(p!=f)DFS(p,u);    int l=g[u].size();    BigInteger cj=1;    for(int p : g[u])        if(p!=f)        {            dp[u][2]*=(dp[p][0]+dp[p][1]+dp[p][2]-dp[p][3]);            //xj*=(dp[p][0]+dp[p][1]+dp[p][2]-dp[p][3]);        }    cj=dp[u][2];    for(int i=0;i<l;i++)        for(int j=i+1;j<l;j++)        {            int x=g[u][i],y=g[u][j];            if(x==f || y==f)continue;            dp[u][0]+=cj/(dp[x][0]+dp[x][1]+dp[x][2]-dp[x][3])/(dp[y][0]+dp[y][1]+dp[y][2]-dp[y][3])*(dp[x][1]+dp[x][2])*(dp[y][1]+dp[y][2]);        }    for(int x : g[u])        if(x!=f)        {            dp[u][3]+=cj/(dp[x][0]+dp[x][1]+dp[x][2]-dp[x][3])*dp[x][2];            dp[u][1]+=cj/(dp[x][0]+dp[x][1]+dp[x][2]-dp[x][3])*(dp[x][2]+dp[x][1]);        }}int main(){    IN;OUT;    scanf("%d",&n);    if(n==1){printf("1\n");return 0;}    for(int i=1;i<n;i++)    {        int x,y;        scanf("%d%d",&x,&y);        g[x].pb(y);g[y].pb(x);    }    DFS(1,0);    cout<<dp[1][0]+dp[1][1]+dp[1][2]-dp[1][3]<<endl;    return 0;}

C Domino in Casino

给1个n*m的矩阵（元素均为正数），你可以选k个不重叠的1*2小矩形，问所以小矩形的2个数的乘积的和最大是多少。
就是简单网络流

#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<functional>#include<iostream>#include<cmath>#include<cctype>#include<ctime>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=Pre[x];p;p=Next[p])#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  #define Lson (x<<1)#define Rson ((x<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (100000007)#define MAXN (100000+10)#define MAXM (100000+10)long long mul(long long a,long long b){return (a*b)%F;}long long add(long long a,long long b){return (a+b)%F;}long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}typedef long long ll;class Cost_Flow  {  public:      int n,s,t;      int q[MAXM];      int edge[MAXM],Next[MAXM],Pre[MAXN],weight[MAXM],size;      int cost[MAXM];      void addedge(int u,int v,int w,int c)        {            edge[++size]=v;            weight[size]=w;            cost[size]=c;            Next[size]=Pre[u];            Pre[u]=size;        }        void addedge2(int u,int v,int w,int c=0){addedge(u,v,w,c),addedge(v,u,0,-c);}       bool b[MAXN];      int d[MAXN];      int pr[MAXN],ed[MAXN];      bool SPFA(int s,int t)        {            For(i,n) d[i]=INF,b[i]=0;         d[q[1]=s]=0;b[s]=1;            int head=1,tail=1;            while (head<=tail)            {                int now=q[head++];                Forp(now)                {                    int &v=edge[p];                    if (weight[p]&&d[now]+cost[p]<d[v])                    {                        d[v]=d[now]+cost[p];                        if (!b[v]) b[v]=1,q[++tail]=v;                        pr[v]=now,ed[v]=p;                    }                }                b[now]=0;            }            return d[t]!=INF;        }       int totcost;        int CostFlow(int s,int t)        {            int maxflow=0;        while (SPFA(s,t))            {                int flow=INF;                for(int x=t;x^s;x=pr[x]) flow=min(flow,weight[ed[x]]);             totcost+=flow*d[t];             maxflow+=flow;               for(int x=t;x^s;x=pr[x]) weight[ed[x]]-=flow,weight[ed[x]^1]+=flow;                 }    //        cout<<maxflow<<endl;        return totcost;        }        void mem(int n,int t)      {          (*this).n=n;          size=1;          totcost=0;          MEM(Pre) MEM(Next)       }  }S;  int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} ll id[120][120];ll a[120][120];int main(){    freopen("domino.in","r",stdin);    freopen("domino.out","w",stdout);    int m=read(),n=read(),k=read();    int p1=0,p2=0;    For(i,m) For(j,n) {        a[i][j]=read();        id[i][j]=++p1;    }    int s=++p1,ta=++p1,tb=++p1;    S.mem(p1,p1);    For(i,m) For(j,n) {        if ((i+j)%2==0) {            S.addedge2(s,id[i][j],1);            if (i<m) S.addedge2(id[i][j],id[i+1][j],1,-a[i][j]*a[i+1][j]);            if (j<n) S.addedge2(id[i][j],id[i][j+1],1,-a[i][j]*a[i][j+1]);            if (i>1) S.addedge2(id[i][j],id[i-1][j],1,-a[i][j]*a[i-1][j]);            if (j>1) S.addedge2(id[i][j],id[i][j-1],1,-a[i][j]*a[i][j-1]);        }else {            S.addedge2(id[i][j],ta,1);        }    }    S.addedge2(ta,tb,k);    cout<<-S.CostFlow(s,tb)<<endl;    return 0;}

E Set Partitions

我们定义一个集合大小的比较方式是，集合中的元素从小到大排序后得到的序列的字典序。定义一个集合的划分的比较是，先比较最小的，再比较次小的……，直到比出为止。现在有1~n个整数拆分到若干非空集合中，已知某个划分方案，问这个划分的下一个划分（next_permutation）是？

#include<cstdio>#include<set>using namespace std;int a[501];int main(){    freopen("partitions.in","r",stdin);    freopen("partitions.out","w",stdout);    int n,m;    while(scanf("%d%d",&n,&m)==2&&n){        int cnt=0;        getchar();        for(int i=0;i<m;i++){            do scanf("%d",&a[cnt++]);            while(getchar()!='\n');            a[cnt++]=0;        }        set<int> s;s.insert(a[cnt-2]);        bool flag=false;        for(int i=cnt-3;i>0;i--){            if(!a[i])m--;            else s.insert(a[i]);            if(a[i-1]){                auto it=s.upper_bound(a[i]?a[i]:a[i-1]);                if(it!=s.end()){                    a[i]=*it;                    s.erase(it);                    cnt=i+1;a[cnt]=0;                    flag=true;break;                }            }        }        printf("%d ",n);        if(flag){            printf("%d\n",m+s.size());            for(int i=0;i<=cnt;i++){                if(a[i])printf("%d ",a[i]);                else printf("\n");            }            for(int t:s)                printf("%d\n",t);        }        else{            printf("%d\n",n);            for(int i=1;i<=n;i++)                printf("%d\n",i);        }        printf("\n");    }}close

F Pipe Layout

r*c的矩阵的哈密尔顿回路个数（r*c<=100）

经典题，打表过。

I HTML Table

大模拟

#include<bits/stdc++.h> using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,0x3f,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define MEMx(a,b) memset(a,b,sizeof(a));#define INF (0x3f3f3f3f)#define F (1000000007)#define pb push_back#define mp make_pair#define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;#define PRi2D(a,n,m) For(i,n) { \                        For(j,m-1) cout<<a[i][j]<<' ';\                        cout<<a[i][m]<<endl; \                        } #pragma comment(linker, "/STACK:102400000,102400000")#define ALL(x) (x).begin(),(x).end()#define gmax(a,b) a=max(a,b);#define gmin(a,b) a=min(a,b);#define fr(x) freopen(x,"r",stdin)#define fw(x) freopen(x,"w",stdout)typedef long long ll;typedef long double ld;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return ((a-b)%F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}inline int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} string Filename="table";int f[3100][3100]={},color=0;char Map[3100][3100]={};int last_tag;int col,row;bool _is_end;// last_tag = 0(table) 1(tr) 2(td) 3(col) 4(wol) int main(){    freopen((Filename+".in").c_str(), "r", stdin);    freopen((Filename+".out").c_str(), "w", stdout);    MEM(f)    char ch,lch;    int n=0,m=0,cnt=0;    int p=0,l=0,pl[3]={};    int nowi=0,nowj;    lch=0;    bool b=0;    int N=0,M=0;    while(1) {        lch=ch;        ch=getchar();        ch=tolower(ch);//      cout<<lch<<ch;        if (ch=='<') {            b=1;            ++p;            if (p==4) {                For(l,2)                    if (!pl[l]) pl[l]=1;                    ++color;                while(f[nowi][nowj]) ++nowj;                Fork(i,nowi,nowi+pl[1]-1) {                    Fork(j,nowj,nowj+pl[2]-1) {                        if (f[i][j]) {                            puts("ERROR");return 0;                        }                        f[i][j]=color;//                      cout<<i<<' '<<j<<endl;                    }                }                 Map[2*nowi-1][2*nowj-1] ='+';                Map[2*nowi-1][2*nowj+pl[2]*2-1] ='+';                Map[2*nowi+pl[1]*2-1][2*nowj-1] ='+';                Map[2*nowi+pl[1]*2-1][2*nowj+pl[2]*2-1] ='+';                gmax(N,2*nowi+pl[1]*2-1);                gmax(M,2*nowj+pl[2]*2-1);                Fork(i,2*nowi,2*nowi+pl[1]*2-2) {                    if (Map[i][2*nowj-1]!='+')                        Map[i][2*nowj-1]='|';                    if (Map[i][2*nowj+pl[2]*2-1]!='+')                              Map[i][2*nowj+pl[2]*2-1]='|';                }                Fork(j,2*nowj,2*nowj+pl[2]*2-2) {                    if (Map[2*nowi-1][j]!='+')                        Map[2*nowi-1][j]='-';                    if (Map[2*nowi+pl[1]*2-1][j]!='+')                    Map[2*nowi+pl[1]*2-1][j]='-';                }//              cout<<nowi<<' '<<pl[1]<<' '<<nowj<<' '<<(-1+nowj+pl[2])<<endl;                nowj+=pl[2];                pl[1]=pl[2]=l=0;            }        }        if(b && ch=='>') {            b=0;        }        if (!b) continue;        if (ch=='/') {            p-=2;            if (!p) break;        }        if(lch=='t'&&ch=='r' && p==2) {            ++nowi;nowj=1;        }        if (ch=='r') {            l=1;                    } else if (ch=='c') {            l=2;        }        if (isdigit(ch)) {            pl[l]=pl[l]*10+ch-'0';        }        if (ch==EOF) break;    }//  For(i,n) {//      For(j,m) {//          cout<<f[i][j]<<' ';//      }//  }    For(i,N){        int tm=0;        For(j,M) if (Map[i][j]) tm=j;        For(j,tm) {            putchar(Map[i][j]?Map[i][j]:' ');        }cout<<endl;    }    return 0;}