Codevs_1166_[NOIP2007]_矩阵取数游戏_(动态规划+高精度)

描述

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http://codevs.cn/problem/1166/

分析

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 #include <iostream>
 #include <cstring>
 #include <cstdio>
 #include <algorithm>
 using namespace std;

 const int maxn=;

 struct Bign{
     int x[maxn],cnt;
     int & operator [] (int id){ return x[id]; }
     Bign(){ memset(x,,sizeof x); }//空的构造函数
     Bign(int num){ *this=num; }//赋一个数字的构造函数
     Bign(char* num){ *this=num; }//赋一个字符串数字的构造函数
     Bign operator = (const char* num){//字符串的赋值函数
         memset(x,,sizeof x); cnt=strlen(num);
         for(int i=;i<=cnt;i++) x[i]=num[cnt-i-]-'';
         return *this;
     }
     Bign operator = (int num){//数字的赋值函数
         memset(x,,sizeof x); cnt=;
         while(num){
             x[++cnt]=num%;
             num/=;
         }
         return *this;
     }
     Bign operator + (Bign y) const{//加法
         Bign z; z.cnt=max(cnt,y.cnt);
         for(int i=;i<=z.cnt;i++){
             z[i]+=x[i]+y[i];
             if(z[i]>=) z[i+]++, z[i]-=;
         }
         if(z[z.cnt+]) z.cnt++;//加法至多多出一位
         return z;
     }
     Bign operator - (Bign y) const{//减法
         Bign z; z.cnt=cnt;
         for(int i=;i<=z.cnt;i++){
             z[i]+=x[i]-y[i];
             if(z[i]<) z[i+]--, z[i]+=;
         }
         while(z.cnt&&!z[z.cnt]) z.cnt--;//减法可能会少很多位
         return z;
     }
      Bign operator * (Bign y) const{//乘法
         Bign z;
         for(int i=;i<=cnt;i++)
             for(int j=;j<=y.cnt;j++)
                 z[i+j-]+=x[i]*y[j],z[i+j]+=z[i+j-]/, z[i+j-]%=;
         z.cnt=cnt+y.cnt;//乘法后结果的位数为cnt1+cnt2或cnt1+cnt2-1.
         if(!z[z.cnt]) z.cnt--;
         return z;
     }
     Bign operator / (Bign y) const{//除法
         Bign z=*this,d=;
         for(int i=cnt;i;i--){
             d=d*+x[i];
             for(int j=;j<;j++)
                 if(d<y*(j+)){
                     z[i]=j;
                     d=d-y*j;
                     break;
                 }
         }
         while(z.cnt&&!z[z.cnt]) z.cnt--;
         return z;
     }
     Bign operator % (Bign y) const{//取模
         Bign d=;
         for(int i=cnt;i;i--){
             d=d*+x[i];
             for(int j=;j<;j++)
                 if(d<y*(j+)){
                     d=d-y*j;
                     break;
                 }
         }
         return d;
     }
     Bign &operator += (Bign y){ *this=*this+y; return *this; }
     Bign &operator -= (Bign y){ *this=*this-y; return *this; }
     Bign &operator *= (Bign y){ *this=*this*y; return *this; }
     Bign &operator /= (Bign y){ *this=*this/y; return *this; }
     Bign &operator %= (Bign y){ *this=*this%y; return *this; }

     bool operator < (Bign y){
         if(cnt!=y.cnt) return cnt<y.cnt;
         for(int i=cnt;i;i--)
             if(x[i]!=y[i]) return x[i]<y[i];
         return false;
     }
     bool operator > (Bign y){ return y<*this; }
     bool operator <= (Bign y){ return!(y<*this); }
     bool operator >= (Bign y){ return!(*this<y); }
     bool operator != (Bign y){ return y<*this||*this<y; }
     bool operator == (Bign y){ return!(y<*this)&&!(*this<y); }
 }dp[][],a[][],ans;
 istream& operator >> (istream &in,Bign &x){//输入
     char s[maxn];
     scanf("%s",s+); int len=strlen(s+);
     for(int i=len;i;i--)
         x[len+-i]=s[i]-'';
     x.cnt=len;
     return in;
 }
 ostream& operator << (ostream &out,Bign &x){//输出
     for(int i=x.cnt;i;i--)
         out << x[i];
     if(!x.cnt) out << ;
     return out;
 }

 int main(){
     int n,m;
     scanf("%d%d",&n,&m);
     Bign t=;
     for(int i=;i<=m;i++) t*=;
     for(int i=;i<=n;i++){
         memset(dp,,sizeof dp);
         for(int j=;j<=m;j++){
             cin >> a[i][j];
             dp[j][j]=t*a[i][j];
         }
         Bign tmp=t;
         for(int k=;k<=m;k++){
             tmp/=;
             for(int l=;l<=m-k+;l++){
                 int r=l+k-;
                 if(dp[l+][r]+a[i][l]*tmp>dp[l][r-]+a[i][r]*tmp)
                     dp[l][r]=dp[l+][r]+a[i][l]*tmp;
                 else
                     dp[l][r]=dp[l][r-]+a[i][r]*tmp;
             }
         }
         ans+=dp[][m];
     }
     cout << ans <<endl;
     return ;
 }