poj3670

O(n)的dp

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#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;

#define maxn 30005

int n;
int cow[maxn];
int f[maxn][5];

void input()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
        scanf("%d", &cow[i]);
}

void work()
{
    memset(f, 0, sizeof(f));
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= 3; j++)
        {
            f[i][j] = f[i - 1][1];
            for (int k = 2; k <= j; k++)
                f[i][j] = min(f[i][j], f[i - 1][k]);
            if (cow[i] != j)
                f[i][j]++;
//            printf("%d %d %d\n", i, j, f[i][j]);
        }
    }
    int ans = n;
    for (int i = 1; i <= 3; i++)
        ans = min(ans, f[n][i]);
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= 3; j++)
        {
            f[i][j] = f[i - 1][3];
            for (int k = 2; k >= j; k--)
                f[i][j] = min(f[i][j], f[i - 1][k]);
            if (cow[i] != j)
                f[i][j]++;
        }
    }
    for (int i = 1; i <= 3; i++)
        ans = min(ans, f[n][i]);
    printf("%d\n", ans);
}

int main()
{
    //freopen("t.txt", "r", stdin);
    input();
    work();
    return 0;
}
原文地址:https://www.cnblogs.com/rainydays/p/2574864.html