洛谷 P1005 矩阵取数游戏

题目传送门

解题思路:

这道题说是矩阵,实际上每一行之间没联系,所以只需跑N此区间DP就行了.

f[i][j]表示从i~j能获得的最大分数.

本题唯一难度在于要写高精度,没有高精度感觉就是个黄题,本人做的第一道非模板的高精度题.

AC代码:

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

int n,m,g[90];
struct kkk{
    int s[999],len;
    kkk() {
        memset(s,0,sizeof(s));
        len = 0;
    }
}f[90][90],sum[90],ans;

kkk operator * (const kkk &a,const int &b) {
    kkk c;
    c.len = a.len;
    int x = 0;
    for(int i = 1;i <= c.len; i++) {
        c.s[i] = a.s[i] * b + x;
        x = c.s[i] / 10;
        c.s[i] %= 10;
    }
    while(x > 0) {
        c.s[++c.len] = x % 10;
        x /= 10;
    }
    return c;
    
}

kkk max(const kkk &a,const kkk &b) {
    if(a.len > b.len) return a;
    if(b.len > a.len) return b;
    for(int i = a.len;i >= 1; i--) {
        if(a.s[i] > b.s[i]) return a;
        if(b.s[i] > a.s[i]) return b;
    }
    return a;
}

kkk operator + (const kkk &a,const kkk &b) {
    int u = max(a.len,b.len),x = 0;
    kkk c;
    for(int i = 1;i <= u; i++){
        c.s[i] = a.s[i] + b.s[i] + x;
        x = c.s[i] / 10;
        c.s[i] = c.s[i] % 10;
    }
    if(x > 0)
        c.s[++u] += x; 
    c.len = u;
    return c;
} 

inline void wo() {
    sum[0].len = 1;
    sum[0].s[1] = 1;
    for(int i = 1;i <= m + 2; i++)
        sum[i] = sum[i-1] * 2;
}

int main() {
    scanf("%d%d",&n,&m);
    wo();
    while(n--) {
        memset(f,0,sizeof(f));
        for(int i = 1;i <= m; i++)
            scanf("%d",&g[i]);
        for(int i = 1;i <= m; i++)
            for(int j = m;j >= i; j--) {
                f[i][j] = max(f[i][j],f[i-1][j] + sum[m-(j-i+1)] * g[i-1]);
                f[i][j] = max(f[i][j],f[i][j+1] + sum[m-(j-i+1)] * g[j+1]);
            }
        kkk a;
        for(int i = 1;i <= m; i++)
            a = max(a,f[i][i] + sum[m] * g[i]);
        ans = ans + a;
    }
    for(int i = ans.len;i >= 1; i--)
        printf("%d",ans.s[i]);
    return 0;
} 

 //NOIP提高 2007 T3