srm 533

250

---

Description

给你一串数字序列，每次删掉第$i$个数，获得权值$w[i-1]\times w[i+1]$，求最后剩下第一个和最后一个数获得的最大权值和

Solution

简单dp,枚举断点即可

Code

#include <bits/stdc++.h>
using namespace std;
const int N = 55;
typedef long long LL;
LL f[N][N];
vector<int> w;
LL dp(int l, int r) {
    if (r - l == 1) return f[l][r] = 0;
    LL &t = f[l][r];
    if (~t)    return t;
    t = 0;
    for (int i = l + 1; i < r; ++i) {
        t = max(t, dp(l, i) + dp(i, r) + w[l] * w[r]);
    }
    return t;
}
class CasketOfStar {
    public:
    int maxEnergy(vector <int> weight) {
        w = weight;
        int n = weight.size();
        memset(f, -1, sizeof(f));
        return dp(0, n - 1);
    }
};

500

---

Description：

给定一个字符矩阵，我们要按顺序取遍所有的”#”，规定操作顺序是取同行的一个”#”，再取同列的一个”#”，以此类推，问能不能取遍

Solution

很容易反应到，这是一道欧拉回路的题，行列单独考虑，处理一些细节就可以了。注意到如果是欧拉通路的话起始点的问题就可以了。总体来说比较简单。

Code:

#include <bits/stdc++.h>
using namespace std;
const int N = 55;
int r[N], c[N];
bool ok[N][N], vis[N][N];
int n, m, tot;
void dfs(int x, int y) {
    vis[x][y] = 1;
    --tot;
    for (int j = 0; j < m; ++j)
        if (ok[x][j] && !vis[x][j]) dfs(x, j);
    for (int i = 0; i < n; ++i)
        if (ok[i][y] && !vis[i][y]) dfs(i, y);
}
class MagicBoard {
    public:
    string ableToUnlock(vector <string> board) {
        n = board.size(), m = board[0].size();
        int x, y;
        for (int i = 0; i < n; ++i)
            for (int j = 0; j < m; ++j) 
                if (board[i][j] == '#') {
                    ok[i][j] = 1;
                    ++r[i], ++c[j];
                    x = i, y = j;
                    ++tot;
                }
        int rr = 0, cc = 0;
        for (int i = 0; i < n; ++i) if (r[i] & 1)   ++rr;
        for (int j = 0; j < m; ++j) if (c[j] & 1)   ++cc;
        if (rr + cc > 2)    return "NO";
        if (rr + cc == 2 && !cc)    return "NO";
        dfs(x, y);
        cout << tot << endl;
        return tot ? "NO" : "YES";
    }
}P;
int main() {
    vector<string> s;
    s.push_back("##"), s.push_back(".#");
    P.ableToUnlock(s);
    return 0;
}