srm 527

275

题意：

构造出一棵树，给出每种度数的权值，一棵树的权值是 所有点权和，求最大权值

Solution:

我们可以发现由于是一棵树，所有度数之和一定是2n?2，我们要做的其实就是度数的一个划分，使得权值最大而已，dp一下即可。

Code:

#include <bits/stdc++.h>
using namespace std;
const int N = 55;
int dp[N][N << 1];
class P8XGraphBuilder {
    public:
    int solve(vector <int> scores) {
        int n = scores.size() + 1;
        memset(dp, -1, sizeof(dp));
        int m = 2 * n - 2;
        dp[0][0] = 0;
        for (int i = 0; i < n; ++i)
            for (int j = 0; j < m; ++j)
                for (int k = 1; k + j <= m; ++k)
                    if (~dp[i][j]) dp[i + 1][j + k] = max(dp[i + 1][j + k], dp[i][j] + scores[k - 1]);
        return dp[n][m];
    }
};

450

题意：

给你一个0/1矩阵，包含’?’,’?’可以变成0或1，给定每行的情况以及每列的情况，让你输出字典序最小的矩阵

Solution

这个题很裸，显然可以看出是二分图匹配，枚举问号，先枚举为0，跑匹配看是否为最大匹配，否则为1即可。

Code:

#include <bits/stdc++.h>
using namespace std;
const int N = 35;
int n, m, g[N][N], l[N], v[N];
bool check(vector <string> a, vector <string> b, int x, int y) {
    for (int i = 0; i < n; ++i) {
        char c = a[i][x], d = b[y][i];
        if (c != d && c != ‘?‘ && d != ‘?‘) return 0;
    }
    return 1;
}
bool can(int u) {
    for (int i = 0; i < m; ++i) {
        if (g[u][i] && !v[i]) {
            v[i] = 1;
            if (l[i] == -1 || can(l[i])) {
                l[i] = u;
                return 1;
            }
        }
    }
    return 0;
}
bool ok() {
    memset(l, -1, sizeof(l));
    int cnt = 0;
    for (int i = 0; i < m; ++i) {
        memset(v, 0, sizeof(v));
        if (can(i)) ++cnt;
    }
    return cnt == m;
}
class P8XMatrixRecovery {
    public:
    vector <string> solve(vector <string> rows, vector <string> columns) {
        n = rows.size(), m = columns.size();
        for (int i = 0; i < n; ++i)
            for (int j = 0; j < m; ++j) {
                if (rows[i][j] != ‘?‘)  continue;
                rows[i][j] = ‘0‘;
                memset(g, 0, sizeof(g));
                for (int k = 0; k < m; ++k)
                    for (int l = 0; l < m; ++l)
                        if (check(rows, columns, k, l)) g[k][l] = 1;
                if (!ok())  rows[i][j] = ‘1‘;
            }
        return rows;
    }
};