cf dp 题

D. Armchairs

#include<bits/stdc++.h>
using namespace std;
int n;
const int N = 5003;
int a[N]; // 1
int b[N]; // 0
int dp[N][N];
int main()
{
	int x; cin >> n;
	int pa = 0, pb = 0;
	for(int i = 1; i <= n; ++i)
	{
		scanf("%d", &x);
		if(x) a[++pa] = i;
		else b[++pb] = i;
	}
	memset(dp, 0x3f, sizeof(dp));
	for(int i = 0; i <= pb; ++i) dp[0][i] = 0;
	for(int i = 1; i <= pa; ++i)
		for(int j = 1; j <= pb; ++j)
			dp[i][j] = min(dp[i][j-1], dp[i-1][j-1] + abs(a[i] - b[j]));
	cout << dp[pa][pb] << endl;
	return 0;
}

待补 CD dp题