题目链接
贪心模拟了半天，最后放弃了
题意
给你一串从$1-n$的序列，其中部分未知（表示为0），补全序列使得相邻数值奇偶性相反的数量最少
相邻数值的奇偶性相反：两个相邻的两个数值，其中一个为奇数另外一个为偶数
分析
一开始用了贪心，结果卡在第十二个样例，然后改成dp
定义dp数组如下
得到状态转移方程
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| dp[i][j][1] = min(dp[i - 1][j - 1][0] + 1, dp[i - 1][j - 1][1]);
dp[i][j][0] = min(dp[i - 1][j][1] + 1, dp[i - 1][j][0]);
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当然这得看这个位置本身是不是已经有了数值，如果为0则两个都需要，如果已经有数值了就按照原来的数值进行dp
AC代码
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| #include <bits/stdc++.h>
using namespace std;
void solve() {
int n;
int dp[120][60][2], value[120];
cin >> n;
for (int i = 0; i < n; ++i) {
cin >> value[i];
}
memset(dp, 0x3f, sizeof(dp));
if (value[0] == 0)
dp[0][1][1] = dp[0][0][0] = 0;
else
dp[0][value[0] & 1][value[0] & 1] = 0;
for (int i = 1; i < n; ++i) {
for (int j = 0; j <= min(i + 1, (n + 1) / 2); ++j) {
if ((value[i] & 1 || value[i] == 0) && j > 0)
dp[i][j][1] = min(dp[i - 1][j - 1][0] + 1, dp[i - 1][j - 1][1]);
if (!(value[i] & 1))
dp[i][j][0] = min(dp[i - 1][j][1] + 1, dp[i - 1][j][0]);
}
}
cout << min(dp[n - 1][(n + 1) / 2][1], dp[n - 1][(n + 1) / 2][0]) << endl;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
#ifdef ACM_LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
long long test_index_for_debug = 1;
char acm_local_for_debug;
while (cin >> acm_local_for_debug) {
cin.putback(acm_local_for_debug);
if (test_index_for_debug > 20) {
throw runtime_error("Check the stdin!!!");
}
auto start_clock_for_debug = clock();
solve();
auto end_clock_for_debug = clock();
cout << "Test " << test_index_for_debug << " successful" << endl;
cerr << "Test " << test_index_for_debug++ << " Run Time: "
<< double(end_clock_for_debug - start_clock_for_debug) / CLOCKS_PER_SEC << "s" << endl;
cout << "--------------------------------------------------" << endl;
}
#else
solve();
#endif
return 0;
}
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