POJ 3671 Dining Cows (DP)

To give you a string of only 1, 2 of the number, let you change the minimum number of times, so that the sequence becomes non-descending.

Thinking: Dynamic programming, judging the demarcation point, open an array of dp[30010][2], where dp[i][j] means to change the number of I to J at least how many times it takes
Then the state transfer equation is listed:
Make A=1 j! =a[i]
0 J==a[i]
So dp[i][1]=dp[i-1][1]+a;
Dp[i][2]=min (dp[i-1][1],dp[i-1][2]) +a;

#include <iostream>using namespace Std;int dp[30010][2];int main () {    int n;    cin>>n;    for (int i=1; i<=n; i++)    {        int a, b;        cin>>a;        b= (a==1?0:1);        Dp[i][1]=dp[i-1][1]+b;        B=!b;        Dp[i][2]=min (Dp[i-1][1], dp[i-1][2]) +b;    }    Cout<<min (Dp[n][1], dp[n][2]) <<endl;    return 0;}

POJ 3671 Dining Cows (DP)