POJ 3666 DP

题意：

思路：

dp[i][j] 表示前i + 1个数变成单调且最后一个数是B[j]，此时的最小成本

dp[i][j] = min(dp[i – 1][k]) + |A[i] – B[j]| 【k = 0->j】

但是我们发现现在的复杂度是O(n^3) 卡不过去

怎么优化呢

保存个最小值不就行了嘛….复杂度O(n^2)

Ps：这道题可以优化空间…

//By SiriusRen
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define N 2222
int n,a[N],b[N],c[N],f[N][N],ans=0x7fffffff;
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++)scanf("%d",&a[i]),b[i]=a[i];
    sort(b+1,b+1+n);
    for(int i=1;i<=n;i++)c[i]=lower_bound(b+1,b+1+n,a[i])-b;
    memset(f,0x7f,sizeof(f));
    for(int i=1;i<=n;i++)f[0][i]=0;
    for(int i=1;i<=n;i++){
        int minn=0x7fffffff;
        for(int j=1;j<=n;j++){
            minn=min(minn,f[i-1][j]);
            f[i][j]=minn+abs(b[j]-b[c[i]]);
        }
    }
    for(int i=1;i<=n;i++)ans=min(ans,f[n][i]);
    printf("%d\n",ans);
}

优化空间的版本~

//By SiriusRen
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define N 2222
int n,a[N],b[N],c[N],f[2][N],ans=0x7fffffff;
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++)scanf("%d",&a[i]),b[i]=a[i];
    sort(b+1,b+1+n);
    for(int i=1;i<=n;i++)c[i]=lower_bound(b+1,b+1+n,a[i])-b;
    memset(f,0x7f,sizeof(f));
    for(int i=1;i<=n;i++)f[0][i]=0;
    for(int i=1;i<=n;i++){
        int minn=0x7fffffff;
        for(int j=1;j<=n;j++){
            minn=min(minn,f[(i+1)%2][j]);
            f[i%2][j]=minn+abs(b[j]-b[c[i]]);
        }
    }
    for(int i=1;i<=n;i++)ans=min(ans,f[n%2][i]);
    printf("%d\n",ans);
}