dp[i][j][k]表示前i個數使用j個偶數末尾爲k時的最小結果
#include<cstdio>
#include<algorithm>
using namespace std;
int w[110], n, C0, C1, D[110][110][2];
int main() {
int i, j, k, l;
scanf("%d", &n);
for (i = 1; i <= n; i++) {
scanf("%d", &w[i]);
if (i % 2 == 0)C0++;
else C1++;
}
for (i = 0; i <= n; i++) {
for (j = 0; j <= n; j++) {
D[i][j][0] = D[i][j][1] = 1e9;
}
}
D[0][0][0] = D[0][0][1] = 0;
for (i = 1; i <= n; i++) {
for (j = 0; j < i; j++) {
for (k = 0; k < 2; k++) {
int t = D[i - 1][j][k];
if (w[i]) {
if (w[i] % 2 == 0) {
D[i][j + 1][0] = min(D[i][j + 1][0], t + (k != 0));
}
else {
D[i][j][1] = min(D[i][j][1], t + (k != 1));
}
}
else {
for (l = 0; l < 2; l++) {
D[i][j + 1 - l][l] = min(D[i][j + 1 - l][l], t + (k != l));
}
}
}
}
}
printf("%d\n", min(D[n][C0][0], D[n][C0][1]));
}