tongji 30029 插头dp

http://acm.tongji.edu.cn/problem?pid=30029

Hnoi2004 postman 

题意：从左上角出发，最后回到左上角，要求经过所有点至少一次，求步数最小的方案数。(n<=10,m<=20)

分析：有三种情况。
1.n，m至少有一个为1，则答案为1
2.n，m至少有一个偶数，则答案为哈密顿回路条数*2，
  此时正好经过所有点一次除(1,1)外
3.【数据里并没有出现这种情况】
  n, m均为大于1的奇数，这时不存在哈密顿回路，
  则最短步数应为从(1,1)出发，终于(1,3),(2,2),(3,1)。
  答案为到上面任一种的条数*3
参考http://blog.csdn.net/fp_hzq/article/details/7523056

还有，要保持n>m,因为当列数12已经是极限了，是不能达到20的，所以应该交换。
保证n>m(即m<=10)后速度得到了极大提升。

最后用上高精，因为10 20答案超了long long

给出几组数据

4 4
12
6 6
2144
9 16
619313040592944136
10 20  
177029033285148340652006844

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<cstdlib>#define MS(x, y) memset(x, y, sizeof(x))#define FOR(i, x, y) for(int i=x; i<=y; i++)#define rFOR(i, x, y) for(int i=x; i>=y; i--)using namespace std;const int maxn = 100;struct bigint{   int s[maxn];    bigint() {MS(s, 0);}    bigint(int num) {*this = num;}    bigint(const char* num){*this = num;}    bigint operator = (const char* num) //字符串赋值    {        MS(s, 0);        if(num[0] == '-')        {            num = num + 1;            s[0] = -1;        }        else s[0] = 1;        while(num[0] == '0') num = num + 1;        s[0] = s[0] * strlen(num);        int len = abs(s[0]);        FOR(i, 1, len) s[i] = num[len - i] - 48;        return *this;    }    bigint operator = (int num)         //整数赋值    {        char s[maxn];        sprintf(s, "%d", num);        *this = s;        return *this;    }    string str() const                      //返回字符串    {        string res = "";        FOR(i, 1, abs(s[0])) res = (char)(s[i] + 48) + res;        if(res == "") return res = "0";        if(s[0] < 0) res = '-' + res;        return res;    }    //----------以上是构造部分，以下是运算部分----------//    bool operator < (const bigint& b) const    {        if(s[0] != b.s[0]) return s[0] < b.s[0];        int len = abs(s[0]);        rFOR(i, len, 1)        if(s[i] != b.s[i])            return (s[i] < b.s[i]) ^ (s[0] < 0);        return false;    }    bool operator > (const bigint& b) const{return b < *this;}    bool operator <= (const bigint& b) const{return !(b < *this);}    bool operator >= (const bigint& b) const{return !(*this < b);}    bool operator != (const bigint& b) const{return b < *this || *this < b;}    bool operator == (const bigint& b) const{return !(b < *this) && !(*this < b);}    friend bigint abs(bigint b)    {        b.s[0] = abs(b.s[0]);        return b;    }    friend bigint _add(const bigint& a, const bigint& b)    {        bigint c;        c.s[0] = max(a.s[0], b.s[0]);        FOR(i, 1, c.s[0]) c.s[i] = a.s[i] + b.s[i];        FOR(i, 1, c.s[0])        if(c.s[i] >= 10)        {            c.s[i+1]++;            c.s[i]-=10;        }        if(c.s[c.s[0]+1]) ++c.s[0];        return c;    }    friend bigint _sub(const bigint& a, const bigint& b)    {        bigint c;        c.s[0] = a.s[0];        FOR(i, 1, c.s[0]) c.s[i] = a.s[i] - b.s[i];        FOR(i, 1, c.s[0])        if(c.s[i] < 0)        {            c.s[i+1]--;            c.s[i] += 10;        }        while(!c.s[c.s[0]] && c.s[0]) --c.s[0];        return c;    }    bigint operator + (const bigint& b) const    {        if(s[0] >= 0 && b.s[0] >= 0) return _add(*this, b);        if(b.s[0] < 0) return *this - abs(b);        if(s[0] < 0) return b - abs(*this);    }    bigint operator - (const bigint& b) const    {        if(s[0] >= 0 && b.s[0] >= 0)        {            bigint c;            if(*this >= b) return _sub(*this, b);            c = _sub(b, *this);            c.s[0] = -c.s[0];            return c;        }        if(b.s[0] < 0) return *this + abs(b);        if(s[0] < 0)        {            bigint c;            c = abs(*this) + b;            c.s[0] = -c.s[0];            return c;        }    }    bigint operator * (const bigint& b) const    {        bigint c;        c.s[0] = abs(s[0]) + abs(b.s[0]);        FOR(i, 1, abs(s[0]))        FOR(j, 1, abs(b.s[0]))            c.s[i + j - 1] += s[i] * b.s[j];        FOR(i, 1, c.s[0])        {            c.s[i+1] += c.s[i] / 10;            c.s[i] %= 10;        }        while(!c.s[c.s[0]] && c.s[0]) --c.s[0];        if((s[0] > 0) != (b.s[0] > 0)) c.s[0] = -c.s[0];        return c;    }    bigint operator / (const bigint& _b) const    {        bigint c, t;        bigint b = abs(_b);        c.s[0] = abs(s[0]);        rFOR(i, c.s[0], 1)        {            rFOR(j, t.s[0], 1) t.s[j + 1] = t.s[j];            t.s[1] = s[i];            ++t.s[0];            while(t >= b)            {                ++c.s[i];                t -= b;            }        }        while(!c.s[c.s[0]] && c.s[0]) --c.s[0];        if((s[0] > 0) != (b.s[0] > 0)) c.s[0] = -c.s[0];        return c;    }    bigint operator % (const bigint& _b) const    {        bigint c, t;        bigint b = abs(_b);        rFOR(i, abs(s[0]), 1)        {            rFOR(j, t.s[0], 1) t.s[j + 1] = t.s[j];            t.s[1] = s[i];            ++t.s[0];            while(t >= b) t -= b;        }        if((s[0] < 0)) t.s[0] = -t.s[0];        return t;    }    bigint operator += (const bigint& b) {*this = *this + b;return *this;}    bigint operator -= (const bigint& b) {*this = *this - b;return *this;}    bigint operator *= (const bigint& b) {*this = *this * b;return *this;}    bigint operator /= (const bigint& b) {*this = *this / b;return *this;}    bigint operator %= (const bigint& b) {*this = *this % b;return *this;}    friend bigint operator + (int& a, const bigint& b){return (bigint)a + b;}    friend bigint operator - (int& a, const bigint& b){return (bigint)a - b;}    friend bigint operator * (int& a, const bigint& b){return (bigint)a * b;}    friend bigint operator / (int& a, const bigint& b){return (bigint)a / b;}    friend bigint operator % (int& a, const bigint& b){return (bigint)a % b;}    friend bigint operator <  (int& a, const bigint& b){return (bigint)a < b;}    friend bigint operator <= (int& a, const bigint& b){return (bigint)a <=b;}    friend bigint operator >  (int& a, const bigint& b){return (bigint)a > b;}    friend bigint operator >= (int& a, const bigint& b){return (bigint)a >=b;}    friend bigint operator == (int& a, const bigint& b){return (bigint)a ==b;}    friend bigint operator != (int& a, const bigint& b){return (bigint)a !=b;}};istream& operator >> (istream &in, bigint& x){    string s;    in >> s;    x = s.c_str();    return in;}ostream& operator << (ostream &out, const bigint& x){    out << x.str();    return out;}typedef bigint LL;const int N=22;const int HASH=15111;const int STATE=20000;LL ans;int n,m;int map[N][N];int jz[N];int ex,ey;struct HASHMAP{    int head[HASH],next[STATE],size;    int state[STATE];    LL f[STATE];    void init()    {        size=0;        memset(head,-1,sizeof(head));    }    void push(int st,LL ans)    {        int i;        int h=st%HASH;        for(i=head[h];i!=-1;i=next[i])          if(state[i]==st)          {              f[i]+=ans;              return;          }        state[size]=st;        f[size]=ans;        next[size]=head[h];        head[h]=size++;    }}hm[2];void shift(int cur){    for(int i=0;i<hm[cur].size;i++)     hm[cur].state[i]<<=2;}inline int cp(int p,int i){    return p&(~(3<<jz[i]));}inline int cp(int p,int i,int j){    return p&(~(3<<jz[i]))&(~(3<<jz[j]));}inline int cp(int p,int i,int j,int k){    return p&(~(3<<jz[i]))&(~(3<<jz[j]))&(~(3<<jz[k]));}inline int getp(int p,int i){    return 3&(p>>jz[i]);}inline int pp(int i,int k){   return k<<jz[i];}inline int lp(int p,int j){   for (int it=j-2,cnt=1;;it--)        {   int pi=getp(p,it);            if (pi==1) cnt--;            else if (pi==2) cnt++;            if (cnt==0) return it;        }}inline int rp(int p,int j){   for (int it=j+1,cnt=1;;it++)        {   int pi=getp(p,it);            if (pi==2) cnt--;            else if (pi==1) cnt++;            if (cnt==0) return it;        }}void dp(int i,int j,int cur){    for(int k=0;k<hm[cur].size;k++)        {int s=hm[cur].state[k];         LL data=hm[cur].f[k];         int left=getp(s,j);         int up=getp(s,j+1);         if (left==0&&up==0)            {if (map[i][j+1]&&map[i+1][j])                {                    hm[cur^1].push(s|pp(j,1)|pp(j+1,2),data);                }             continue;            }         if (left==0||up==0)            {   int w=left+up;                int tmp=cp(s,j,j+1);                if (map[i][j+1]) hm[cur^1].push(tmp|pp(j+1,w),data);                if (map[i+1][j]) hm[cur^1].push(tmp|pp(j,w),data);                continue;            }         if (left==up)            {   int it=(left==1)?rp(s,j+1):lp(s,j+1);                hm[cur^1].push(cp(s,j,j+1,it)|pp(it,left),data);                continue;            }         if (left==1&&up==2)            {   if (i==ex&&j==ey) ans+=data;                continue;            }        if (left==2&&up==1)            {   hm[cur^1].push(cp(s,j,j+1),data);                continue;            }        }}void solve1(){    int cur=0;    ans=0;    hm[cur].init();    hm[cur].push(0,1);    for(int i=0;i<n;i++)        {for(int j=0;j<m;j++)            if (map[i][j])            {             hm[cur^1].init();             dp(i,j,cur);             cur^=1;            }         shift(cur);        }    cout<<ans*2<<endl;}void solve2(){    int cur=0;    ans=0;    hm[cur].init();    hm[cur].push(0,1);    for(int i=0;i<n;i++)        {for(int j=0;j<m;j++)            if (map[i][j])            {             hm[cur^1].init();             dp(i,j,cur);             cur^=1;            }         shift(cur);        }    cout<<ans*3<<endl;}void prepared_jinzhi(){    jz[0]=0;    for (int i=1;i<=m;i++)        jz[i]=i<<1;}void init1(){   if (n<m) {int swap=n;n=m;m=swap;}    memset(map,0,sizeof(map));    for (int i=0;i<n;i++)        for (int j=0;j<m;j++)            map[i][j]=1;    ex=n-1; ey=m-1;}void init2(){   if (n<m) {int swap=n;n=m;m=swap;}    memset(map,0,sizeof(map));    n+=2; m+=4;    for (int i=2;i<n;i++)        for (int j=2;j<m-2;j++)            map[i][j]=1;    for (int i=0;i<m;i++)        map[0][i]=1;    map[1][0]=map[1][m-1]=1;    map[2][0]=map[2][1]=map[2][m-2]=map[2][m-1]=1;    ex=n-1; ey=m-3;}int main(){    while(scanf("%d%d",&n,&m)!=EOF)    {   if (n==1||m==1){printf("%d\n",1); continue;}        if (n*m%2==1)            {init2();             prepared_jinzhi();             solve2();            }        else {init1();             prepared_jinzhi();             solve1();             }    }    return 0;}