牛客小白月赛1 G あなたの蛙は旅⽴っています【图存储】【DP】

题目链接：https://www.nowcoder.com/acm/contest/85/G

思路：

DP 空间可以优化成一维的， 用一维数组的 0 号单元保存左斜对角的值即可。

存图这里真不好理解 = =

 AC 代码：

#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <vector>

using namespace std;


#define max3(x, y, z) max(max((x), (y)), (z))
#define min3(x, y, z) min(mix((x), (y)), (z))
#define pb push_back
#define ppb pop_back
#define mk make_pair

#define debug_l(a) cout << #a << " " << (a) << endl
#define debug_b(a) cout << #a << " " << (a) << " "
#define testin(filename) freopen((filename) ,"r",stdin)
#define testout(filename) freopen((filename) ,"w",stdout)

typedef long long ll;
typedef unsigned long long ull;


const double PI  = 3.14159265358979323846264338327;
const double E   = exp(1);
const double eps = 1e-6;

const int INF    = 0x3f3f3f3f;
const int NINF   = 0xc0c0c0c0;
const int maxn   = 3e3 + 5;
const int MOD    = 1e9 + 7;


int Map[maxn][maxn], Cur[maxn], dp[maxn][maxn];
int main()
{
    //testin("data1.in");
    int n;
    scanf("%d", &n);
    memset(Map, NINF, sizeof(Map));
    memset(Cur, 0, sizeof(Cur));
    memset(dp, 0, sizeof(dp));
    int vis = 1 + (4 * (n - 1));
    int cur = 2 * n - 1;
    int i, j;

    for (int i = 0; i < vis - n; i++) {
        if (i <= i % n)
            Cur[i] = i % n + 1;
        else {
            if (n & 1)
                Cur[i] = (i & 1) ? n - 1 : n;
            else
                Cur[i] = (i & 1) ? n : n - 1;
        }
    }
    for (int i = vis - 1; i >= vis - n; i--)
        Cur[i] = vis - i;

    vector <int> v[vis];
    int temp;
    for (i = 0; i < vis; i++)
    {
        for (j = 0; j < Cur[i]; j++)
        {
            scanf("%d", &temp);
            v[i].push_back(temp);
        }
    }



    int len = cur / 2 + 1;
    int flag = 0;
    for (i = 0, j = n; i < len; i++, j++)
    {
        for (int l = 0, k = flag; l < j; l++, k++)
        {
            Map[i][l] = v[k][v[k].size() - 1];
            v[k].pop_back();
            if (v[k].size() == 0)
                flag++;
        }
    }

    for (j -= 2; i < cur; i++, j--)
    {
        for (int l = (cur - j), k = flag; l < cur; l++, k++)
        {
            Map[i][l] = v[k][v[k].size() - 1];
            v[k].pop_back();
            if (v[k].size() == 0)
                flag++;
        }
    }

    dp[0][0] = Map[0][0];
    for (int i = 1; i < cur; i++) {
        dp[0][i] = Map[0][i] + dp[0][i - 1];
        dp[i][0] = Map[i][0] + dp[i - 1][0];
    }

    for (int i = 1; i < cur; i++)
        for (int j = 1; j < cur; j++)
            dp[i][j] = Map[i][j] + max3(dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]);

    cout << dp[cur - 1][cur - 1] << endl;
    return 0;
}

参考原文：https://blog.csdn.net/Dup4plz/article/details/79639771