P1005 矩阵取数游戏（动态规划+高精度）

题目链接：传送门

题目大意：

给定长度为m的数列a_j，每次从两端取一个数，得到2^k * a_j的价值（k为当前的次数，从1开始到m），总共有n行这样的数列，求最大价值总和。

1 ≤ n, m ≤ 80, 0 ≤ a_j ≤ 1000;

思路：

状态f[i][j]表示取剩下a_i，a_i+1，…，a_j时的最大价值。

起始状态：

　　f[0][m-1] = 0；

转移方程：

　　f[i][j-1] = max(f[i][j-1], f[i][j] + 取掉a_j得到的价值)；

　　f[i+1][j] = max(f[i+1][j], f[i][j] + 取掉a_i得到的价值)；

PS：吃灰模板居然出了bug，还因此WA了一发，是时候更新一下高精模板了。

#include <bits/stdc++.h>

using namespace std;
const int maxn = ;
const int MAX_N =  + ;

struct bigInt{
    int len, d[maxn];

    void clean() { while(len >  && !d[len-]) len--; }
    string str() const {
        string s;
        for (int i = ; i < len; i++) s += d[len--i] + '';
        return s;
    }

    bigInt() { memset(d, , sizeof d); len = ; }
    bigInt(int num) { *this = num; }
    bigInt(char* num) { *this = num; }

    bool operator < (const bigInt& b) const {
        if(len != b.len)
            return len < b.len;
        for (int i = len-; i >= ; i--)
            if (d[i] != b.d[i])
                return d[i] < b.d[i];
        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) && !(b > *this);}

    bigInt operator = (const char* num) {
        memset(d, , sizeof d);
        len = strlen(num);
        for (int i = ; i < len; i++)
            d[i] = num[len--i] - '';
        clean();
        return *this;
    }
    bigInt operator = (int num) {
        char s[];
        sprintf(s, "%d", num);
        *this = s;
        return *this;
    }
    bigInt operator + (const bigInt& b) {
        bigInt c = *this;
        for (int i = ; i < b.len; i++) {
            c.d[i] += b.d[i];
            c.d[i+] += c.d[i]/;
            c.d[i] %= ;
        }
        c.len = max(len, b.len)+;
        c.clean();
        return c;
    }
    bigInt operator - (const bigInt& b) {
        bigInt c = *this;
        int i;
        for (i = ; i < b.len; i++) {
            c.d[i] -= b.d[i];
            if (c.d[i] < ) c.d[i] += , c.d[i+]--;
        }
        while (c.d[i] < ) c.d[i++] += , c.d[i]--;
        c.clean();
        return c;
    }//只能正数大减小
    bigInt operator * (const bigInt& b) const {
        bigInt c;
        for (int i = ; i < len; i++)
            for (int j = ; j < b.len; j++)
                c.d[i+j] += d[i] * b.d[j];
        for (int i = ; i < len+b.len || !c.d[i]; c.len = ++i) {
            c.d[i+] += c.d[i] / ;
            c.d[i] %= ;
        }
        c.clean();
        return c;
    }
    bigInt operator / (const bigInt& b) {
        bigInt c = *this, res = ;
        for (int i = ; i < len; i++) {
            res = res* + c.d[len--i];
            int j;
            for (j = ; j < ; j++)
                if(res < b*(j+))
                    break;
            c.d[len--i] = j;
            res = res - b*j;
        }
        c.clean();
        return c;
    }
    bigInt operator % (const bigInt& b) {
        bigInt res = ;
        for (int i = ; i < len; i++) {
            res = res* + d[len--i];
            int j;
            for (j = ; j < ; j++)
                if(res < b*(j+))
                    break;
            res = res - b*j;
        }
        return res;
    }
    bigInt operator += (const bigInt& b) {
        *this = *this + b;
        return *this;
    }
};

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;
}

int N, M;
bigInt mat[MAX_N];
bigInt mul[MAX_N];
bigInt f[MAX_N][MAX_N];

bigInt dp()
{
    bigInt cur;
    for (int i = ; i < M; i++) {
        for (int j = M-; j > i; j--) {
//            cout << "f[" << i << "][" << j << "] = " << f[i][j] << ":" << endl;
//            cout << "f[" << i << "][" << j-1 << "] = " << f[i][j-1] << ' ' << "f[" << i+1 << "][" << j << "] = " << f[i+1][j] << endl;
            f[i][j-] = max(f[i][j-], f[i][j] + mul[M-(j-i+)+]*mat[j]);
            f[i+][j] = max(f[i+][j], f[i][j] + mul[M-(j-i+)+]*mat[i]);
//            cout << "f[" << i << "][" << j-1 << "] = " << f[i][j-1] << ' ' << "f[" << i+1 << "][" << j << "] = " << f[i+1][j] << endl;
        }
    }
    for (int i = ; i < M; i++) {
        cur = max(cur, f[i][i] + mul[M]*mat[i]);
    }
//    cout << cur << endl;
    return cur;
}

int main()
{
    cin >> N >> M;
    mul[] = ;
    for (int i = ; i < MAX_N; i++)
        mul[i] = mul[i-] * ;
    bigInt ans;
    while (N--) {
        for (int i = ; i < M; i++) {
            cin >> mat[i];
        }
        for (int i = ; i < M; i++)
            for (int j = i; j < M; j++)
                f[i][j] = ;
        ans = ans + dp();
    }
    //模板居然出了bug，怕是在硬盘里吃灰吃多了
    while (ans.d[ans.len-] > ) {
        ans.d[ans.len] = ans.d[ans.len-] / ;
        ans.d[ans.len-] %= ;
        ans.len++;
    }
    cout << ans << endl;
    return ;
}